Note: This answer is aimed to provide a complete solution to this challenging problem. It is based upon a series of papers about compositae of functions by Vladimir Kruchinin and a paper about Bell polynomials by Warren P. Johnsen (see references at the end).
If this somewhat extensive elaboration is too cumbersome for a detailed reading, I suggest to read the overview and take instead a look at Manipulation of Bell Polynomials. This was my starter in applying the techniques used here. Another way to go through this answer is to focus on highlighted regions and skip this way less important details.
Note: [2015-03-08]
It's a great pleasure to provide now a complete answer to OPs challenging question. I could close the gap by adding Step 2 and Step 3 of Part 3 of my answer. It was a really tough job to perform all the calculations (and much more verification steps behind the scene). So, if the interested reader is willing to review parts of this answer any fruitful hints to reduce the size of this proof are appreciated.
Overview:
The answer is divided into three parts.
Part 1: Convenient representation of OPs formula
We transform OPs expression conveniently and replace the convolution product $a\diamond b$ with corresponding partial Bell polynomials to ease further calculations. We show
OPs formula is equivalent to (argument $x$ omitted):
\begin{align*}
\sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln\circ g)}=
\sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{1}\\
&+\sum_{l=1}^{n-1}\sum_{m=1}^{l}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{n}{l}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{l,m}^{f}B_{n-l,k}^{g}\qquad\qquad n\geq 1
\end{align*}
Part 2: Derivation of $B_{n,k}^{f\cdot (\ln\circ g)}$
The strategy to proof OPs formula (1) is to derive the partial Bell polynomials for $B_{n,k}^{f\cdot (\ln\circ g)}$ from scratch and show that the so resulting formula is the same as (1). We do so by starting with partial Bell polynomials of $f$ and $g$ and building stepwise more complex partial Bell polynomials.
2.1 Preliminaries:
We will recognise that deriving Bell polynomials is strongly related to composites of generating functions and their Taylor expansion series. We introduce the necessary concepts and notation.
2.2 From $B_{n,k}^{\ln}$ and $B_{n,k}^g$ to $B_{n,k}^{\ln \circ g}$
We create the partial Bell polynomials $B_{n,k}^{\ln \circ g}$ from $B_{n,k}^{\ln}$ and $B_{n,k}^g$ and show
The following is valid
\begin{align*}
B^{\ln\circ g}_{n,k}&=\sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\qquad 1\leq k \leq n\tag{2}
\end{align*}
with $\begin{bmatrix}j\\k\end{bmatrix}$ the unsigned Stirling numbers of the first kind.
2.3 From $B_{n,k}^{f}$ and $B_{n,k}^h$ to $B_{n,k}^{f\cdot h}$
We create the partial Bell polynomials $B_{n,k}^{f \cdot h}$ from $B_{n,k}^{f}$ and $B_{n,k}^h$ and show
The following is valid
\begin{align*}
B_{n,k}^{f\cdot h}&=h^k B_{n,k}^{f} + f^k B_{n,k}^{h}\\
&\qquad+\frac{1}{k!}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}! m_{2}!f^{k-m_1}h^{k-m_2}\tag{3}\\
&\qquad\qquad\cdot \sum_{l=1}^{n-1}\binom{n}{l}B_{l,m_1}^fB_{n-l,m_2}^{h}\qquad\qquad 1\leq k \leq n\\
\end{align*}
We thereby find some nice combinatorial identities and use interesting techniques (by Egorychev) in order to prove them.
2.4 From $B_{n,k}^{f \cdot h}$ and $B_{n,k}^{\ln \circ g}$ to $B_{n,k}^{f\cdot (\ln \circ g)}$
We combine the partial Bell polynomials $B_{n,k}^{f \cdot h}$ and $B_{n,k}^{\ln \circ g}$ and summing from $1 \leq k \leq n$ results in
\begin{align*}
\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)}
&=\sum_{k=1}^{n}\left((\ln \circ g)^k B_{n,k}^{f} + f^k \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\right)\\
&\qquad+\sum_{k=1}^{n}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}f^{k-m_1} (\ln \circ g)^{k-m_2}\\
&\qquad\qquad\cdot\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\frac{(-1)^{j-m_2}}{g^j}\begin{bmatrix}j\\m_2\end{bmatrix}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\tag{4}
\end{align*}
Part 3:
We have now derived an identity for $\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)}$. We finally show that this identity (4) is equivalent to OPs formula (1) and the proof is completed.
Solution:
Part 1:
First we take a closer look at the structure of OPs expression. In order to do so we introduce the abbreviation:
$$B_{n,k}^{f}(x) := B_{n,k}(f^\prime(x),f^{\prime\prime}(x),\ldots,f^{(n-k+1)})$$
OPs expression can therefore be written as
\begin{align*}
\sum_{k=1}^{n}&\ln^{k}(g(x))B_{n,k}^f(x)=\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}(x)\\
&-\sum_{k=1}^{n}\sum_{m=0}^{n-k}\sum_{j=0}^{m}\binom{m}{j}
\frac{\ln^{m-j}(g(x))}{g(x)^k}\frac{d^j}{d(f(x))^j}[(f(x))_k]B_{{n,m+k}_{(f\rightarrow g)^k}}^f(x)
\end{align*}
We see partial Bell polynomials with different complexity. The most complex is $B_{n,k}^{f\cdot (\ln \circ g)}$ which is based upon a composition of $\ln$ with a function $g$ and a multiplication with a function $f$. Next we have $B_{{n,m+k}_{(f\rightarrow g)^k}}^f$, which is an expression containing the Bell polynomials $B_{n,k}^f$ and $B_{n,k}^g$ and is therefore based upon $f$ and $g$. The simplest one is $B_{n,k}^{f}$ based upon $f$ only. Now we reorganise the expression and put thereby $B_{n,k}^{f\cdot (\ln \circ g)}$ on the LHS.
In order to keep the complicated expressions better manageable, we will also sometimes omit the variable $x$. OPs expression can now be written as:
\begin{align*}
\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}=\sum_{k=1}^{n}&(\ln\circ g)^{k}B_{n,k}^f\tag{5}\\
&+\sum_{k=1}^{n}\sum_{m=0}^{n-k}\sum_{j=0}^{m}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{{n,m+k}_{(f\rightarrow g)^k}}^f\qquad\qquad n\geq 1
\end{align*}
The next step is to take a closer look at the convolution product
\begin{align*}
B_{{n,k}_{(f\rightarrow g)^c}}^f:= \frac{(a^{(k-c)\diamond} \diamond b^{c\diamond})_n}{(k-c)!c!}\qquad 1\leq k\leq n, 0\leq c\leq k
\end{align*}
with $a=(f^{\prime},f^{\prime\prime},\ldots)$ and $b=(g^{\prime},g^{\prime\prime},\ldots)$
Note: You might have a look at my first answer to this question. See the section: The convolution product $x \diamond y$ demystified which provides some information about it.
We note that the partial Bell polynomial $B_{n,k}^f$ is defined as:
\begin{align*}
\frac{1}{n!}B_{n,k}^f=[t^n]\frac{1}{k!}\left(\sum_{m\geq 1}f^{(m)}\frac{t^m}{m!}\right)^k\qquad 1 \leq k \leq n
\end{align*}
You might also have a look at the section: Definition of partial Bell polynomials $B_{n,k}$ of my first answer for more information.
With this information at hand we can show that
the following is valid for $1\leq k \leq n$:
\begin{equation*}
B_{{n,k}_{(f\rightarrow g)^c}}^f=
\begin{cases}
B_{n,k}^f\qquad& c=0\\
\sum_{l=1}^{n-1}\binom{n}{l}B_{l,k-c}^fB_{n-l,c}^g\qquad& n>1, 0 < c < k\tag{6}\\
B_{n,k}^g\qquad& c=k\\
\end{cases}
\end{equation*}
We observe for $n>1,0<c<k$
\begin{align*}
\frac{1}{n!}B_{{n,k}_{(f\rightarrow g)^c}}^f&=[t^n]\frac{1}{(k-c)!}\left(\sum_{m\geq 1}f^{(m)}\frac{t^m}{m!}\right)^{k-c}
\frac{1}{c!}\left(\sum_{m\geq 1}g^{(m)}\frac{t^m}{m!}\right)^{c}\\
&=\sum_{l=1}^{n-1}[t^l]\frac{1}{(k-c)!}\left(\sum_{m\geq 1}f^{(m)}\frac{t^m}{m!}\right)^{k-c}
[t^{n-l}]\frac{1}{c!}\left(\sum_{m\geq 1}g^{(m)}\frac{t^m}{m!}\right)^{c}\\
&=\sum_{l=1}^{n-1}\frac{1}{l!}B_{l,k-c}^{f}\frac{1}{(n-l)!}B_{n-l,c}^{g}\\
&=\frac{1}{n!}\sum_{l=1}^{n-1}\binom{n}{l}B_{l,k-c}^{f}B_{n-l,c}^{g}\\
\end{align*}
The cases $n=1$ and $c\in \{0,k\}$ can be easily verified and (6) follows.
With the help of (6) we can now write OPs expression (5) as:
\begin{align*}
\sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln\circ g)}=
\sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{7}\\
&+\sum_{k=1}^{n}\sum_{m=1}^{n-k}\sum_{j=0}^{m}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]\sum_{l=1}^{n-1}\binom{n}{l}B_{l,m}^{f}B_{n-l,k}^{g}\qquad\qquad n\geq 1
\end{align*}
Observe that in (7) the index range of $m$ starts with $1$ instead of $0$. The case $m=0$ is now treated as $\sum_{k=1}^{n}\frac{(f)_k}{g^k}B_{n,k}^g$ corresponding to the case $c=k$ in (6) whereas the other summand $\sum_{k=1}^{n}(\ln\circ g)^{k}B_{n,k}^f$ corresponds to the case $c=0$.
We note that $B_{l,m}^{f}=0$ for $m>l$ and $B_{n-l,k}^{g}=0$ for $k>n-l$. In order to analyse the sum in (7) conveniently we now restrict the index range precisely to those values, which give a contribution to the sum.
Relevant for these considerations is the expression
\begin{align*}
\sum_{k=1}^{n-1}&\sum_{l=1}^{n-1}\sum_{m=1}^{n-k}B_{l,m}^fB_{n-l,k}^g\tag{8}\\
&=\sum_{k=1}^{n-1}\sum_{l=1}^{n-k}\sum_{m=1}^{n-k}B_{l,m}^fB_{n-l,k}^g\tag{9}\\
&=\sum_{k=1}^{n-1}\sum_{l=1}^{n-k}\sum_{m=1}^{l}B_{l,m}^fB_{n-l,k}^g\tag{10}\\
&=\sum_{l=1}^{n-1}\sum_{k=1}^{n-l}\sum_{m=1}^{l}B_{l,m}^fB_{n-l,k}^g\tag{11}\\
\end{align*}
Comment:
(8) Index $k\leq n-1$ since $(l\geq 1$ and $k=n)\Rightarrow (B_{n-l,k}^g=0)$
(9) Index $l\leq n-k$ since otherwise $B_{n-l,k}^g=0$.
(10) Index $m\leq l$ since otherwise $B_{l,m}^g=0$.
(11) Exchange sums with index $l$ and $k$
OPs expression can therefore be written as:
\begin{align*}
\sum_{k=1}^{n}&B_{n,k}^{f\cdot (\ln\circ g)}=
\sum_{k=1}^{n}\left((\ln\circ g)^{k}B_{n,k}^f+\frac{(f)_k}{g^k}B_{n,k}^g\right)\tag{12}\\
&+\sum_{l=1}^{n-1}\sum_{m=1}^{l}\sum_{k=1}^{n-l}\sum_{j=0}^{m}\binom{n}{l}\binom{m}{j}
\frac{(\ln\circ g)^{m-j}}{g^k}\frac{d^j}{d(f)^j}[(f)_k]B_{l,m}^{f}B_{n-l,k}^{g}\qquad\qquad n\geq 1
\end{align*}
Summary of part 1: We have now finalised this part. OPs expression in the form (12) is a convenient representation for further analysis.
And now for something completely different: In the following part 2 we develop a formula for
$$\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}$$
from scratch. In the last part 3 we will show that both formulae are equal and the job is done.
Part 2:
The development of a formula for $\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln\circ g)}$ from scratch is done in several steps. We start with some
Part 2.1: Preliminaries
Let's consider a function $f=f(z)$ and its Taylor series expansion at a point $x$
\begin{align*}
f(z+x)=\sum_{n\geq 0}\frac{f^{(n)}(x)}{n!}z^n
\end{align*}
We will take a look at the $k$-th power of the difference $f(x+z)-f(x)$ since there is a strong relationship with the partial Bell Polynomials $B_{n,k}\ \ (1\leq k\leq n)$:
\begin{align*}
B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)})
&=\frac{1}{k!}\sum_{\pi_k \in C_n}\binom{n}{\lambda_1,\lambda_2,\ldots,\lambda_k}f^{(\lambda_1)}f^{(\lambda_2)}\cdot\ldots\cdot f^{(\lambda_n)}\qquad\\
&=\frac{n!}{k!}\sum_{\pi_k \in C_n}\frac{f^{(\lambda_1)}(x)}{\lambda_1!}\frac{f^{(\lambda_2)}(x)}{\lambda_2!}\cdot\ldots\cdot\frac{f^{(\lambda_n)}(x)}{\lambda_n!}
\end{align*}
Here $f^{(n)}$ is the $n$-th derivative of $f=f(x)$, $C_n$ is the set of compositions
of $n$ and $\pi_k$ is the composition of $n$ with $k$ parts exactly $\lambda_1+\lambda_2+\ldots+\lambda_k=n$.
In order to do so its convenient to define
\begin{align*}
Y_f(x,z) := f(x+z)-f(x) =\sum_{n> 0}\frac{f^{(n)}(x)}{n!}z^n
\end{align*}
and consider the $k$-th power of $Y_f(x,z)$. We get
\begin{align*}
\left(Y_f(x,z)\right)^k=\left(f(x+z)-f(x)\right)^k=\sum_{n\geq k}Y_f^{\triangle}(n,k,x)z^n
\end{align*}
$Y_f^{\triangle}(n,k,x)$ is introduced to denote the coefficient of $z^n$ of $\left(Y_f(x,z)\right)^k$. In the following we use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a power series. We observe
\begin{align*}
Y_f^\triangle(n,k,x)&=[z^n]\left(f(x+z)-f(x)\right)^k\\
&=\sum_{\pi_{k}\in C_n}\frac{f^{(\lambda_1)}(x)}{\lambda_1!}\frac{f^{(\lambda_2)}(x)}{\lambda_2!}\cdot\ldots\cdot\frac{f^{(\lambda_n)}(x)}{\lambda_n!}\\
&=\frac{k!}{n!}B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)})\qquad\qquad 1\leq k \leq n
\end{align*}
It follows
\begin{align*}
B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)})=\frac{n!}{k!}Y_f^\triangle(n,k,x)
\end{align*}
We again use following abbreviation for the $B_{n,k}$:
\begin{align*}
B^{f}_{n,k}(x):=B_{n,k}(f^{\prime},f^{\prime\prime},\ldots,f^{(n-k+1)})
\end{align*}
Part 2.2: Calculation of $B^{\ln \circ g}_{n,k}$
Using the notation introduced in part 2.1 we consider the difference of the composita of $(\ln \circ g)$:
$$Y_{\ln \circ g}(x,z)=\ln(g(x+z))-\ln(g(x))=\sum_{n>0}\frac{(\ln \circ g)^{(n)}(x)}{n!}z^n$$
and calculate the coefficient $Y^\triangle_{\ln \circ g}(n,k,x)$ of $z^n$ of the $k$-th powers
of the composita $Y^{\ln \circ g}(x,z), k\geq 1$.
We do it in three steps. First we calculate $B^{g}_{n,k}$, then $B^{\ln}_{n,k}$ and finally we get $B^{\ln \circ g}_{n,k}$.
Step 2.2.1: Calculation of $B^{g}_{n,k}$
\begin{align*}
Y_g(x,z)&=g(x+z)-g(x)=\sum_{n>0}\frac{g^{(n)}(x)}{n!}z^n\\
\left(Y_g(x,z)\right)^k&=\left(g(x+z)-g(x)\right)^k=\sum_{n\geq k}Y_g^{\triangle}(n,k,x)z^n\\
\\
Y_g^{\triangle}(n,k,x)&=[z^n]\left(g(x+z)-g(x)\right)^k\\
&=\frac{k!}{n!}B^g_{n,k}(x)
\end{align*}
It follows
\begin{align*}
B^g_{n,k}(x)=\frac{n!}{k!}Y_g^{\triangle}(n,k,x)
\end{align*}
Step 2.2.2: Calculation of $B^{\ln}_{n,k}$
\begin{align*}
Y_{\ln}&=\ln(x+z)-\ln(x)=\sum_{n>0}\frac{\ln^{(n)}(x)}{n!}z^n\\
\left(Y_{\ln}\right)^k&=\left(\ln(x+z)-\ln(x)\right)^k=\sum_{n\geq k}Y_{\ln}^{\triangle}(n,k,x)z^n\\
\\
Y_{\ln}^{\triangle}(n,k,x)&=[z^n]\left(\ln(x+z)-\ln(x)\right)^k\\
&=[z^n]\left(\ln\left(1+\frac{z}{x}\right)\right)^k\\
&=[z^n]\sum_{n\geq k}k!(-1)^{n-k}
\begin{bmatrix}n\\k\end{bmatrix}
x^{-n}\frac{z^n}{n!}\tag{13}\\
&=\frac{k!}{n!}(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}x^{-n}\\
&=\frac{k!}{n!}B^{\ln}_{n,k}(x)
\end{align*}
It follows
\begin{align*}
B^{\ln}_{n,k}(x)=
(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}
x^{-n}
\end{align*}
In (13) we use the unsigned Stirling numbers of the first kind $\begin{bmatrix}n\\k\end{bmatrix}$, which are defined as coefficients in the exponential power series for the $k$-th powers of the logarithm $\ln(1+z)$:
$$\sum_{n\geq k}(-1)^{n-k}\begin{bmatrix}n\\k\end{bmatrix}\frac{z^n}{n!}=\frac{1}{k!}\left(\ln (1+z)\right)^k$$
Step 2.2.3: Calculation of $B^{\ln\circ g}_{n,k}$
\begin{align*}
Y_{\ln\circ g}(x,z)&=\ln\left(g(x+z)\right)-\ln\left(g(x)\right)=\sum_{n>0}\frac{(\ln\circ g)^{(n)}(x)}{n!}z^n\\
\left(Y_{\ln\circ g}(x,z)\right)^k&=\left(\ln\left(g(x+z)\right)-\ln\left(g(x)\right)\right)^k=\sum_{n\geq k}Y_{\ln\circ g}^{\triangle}(n,k,x)z^n\\
\end{align*}
According to Theorem 6 in Kruch_3 the composita of $(\ln \circ g)$ are given as
\begin{align*}
Y_{\ln\circ g}^{\triangle}(n,k,x)=\sum_{j=k}^{n}Y_{g}^{\triangle}(n,j,x)\cdot Y_{\ln}^{\triangle}(j,k,g(x))
\end{align*}
It follows
\begin{align*}
\frac{k!}{n!}B^{\ln\circ g}_{n,k}(x)&=\sum_{j=k}^{n}\frac{j!}{n!}B^{g}_{n,j}(x)
\cdot\frac{k!}{j!}B^{\ln}_{j,k}(g(x))\\
&=\sum_{j=k}^{n}\frac{k!}{n!}B^{g}_{n,j}(x)(-1)^{j-k}\begin{bmatrix}j\\k\end{bmatrix}\left(g(x)\right)^{-j}
\end{align*}
We therefore get
\begin{align*}
B^{\ln\circ g}_{n,k}(x)&=\sum_{j=k}^{n}B^{g}_{n,j}(x)(-1)^{j-k}\begin{bmatrix}j\\k\end{bmatrix}\left(g(x)\right)^{-j}\tag{14}
\end{align*}
The next step is to calculate $B_{n,k}$ for the multiplication of $f\cdot h$ of two functions $f$ and $h$.
Part 2.3: Calculation of $B^{f\cdot h}_{n,k}$
We start similarly as in part 2.2:
Step 2.3.1: Derivation of $B^{f\cdot h}_{n,k}$
\begin{align*}
Y_{f\cdot h}(x,z)&=f(x+z)h(x+z)-f(x)h(x)=\sum_{n>0}\frac{(f\cdot h)^{(n)}(x)}{n!}z^n\\
\left(Y_{f\cdot h}(x,z)\right)^k&=\left(f(x+z)h(x+z)-f(x)h(x)\right)^k=\sum_{n\geq k}Y_{f\cdot h}^{\triangle}(n,k,x)z^n\\
\end{align*}
\begin{align*}
Y_{f\cdot h}^{\triangle}(n,k,x)&=[z^n]\left(f(x+z)h(x+z)-f(x)h(x)\right)^k\\
&=[z^n]\sum_{j=0}^{k}\binom{k}{j}\left(f(x+z)h(x+z)\right)^j\left(-f(x)h(x)\right)^{k-j}\\
&=\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)h(x)\right)^{k-j}[z^n]\left(f(x+z)h(x+z)\right)^j\\
&=\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)h(x)\right)^{k-j}\\
&\qquad\cdot\sum_{i=0}^{n}\left([z^i]\left(f(x+z)\right)^j\right)\left([z^{n-i}]\left(h(x+z)\right)^j\right)\tag{15}
\end{align*}
We want to replace the expressions $[z^n]\left(f(x+z)\right)^j$ in (15) with an expression containing $Y_{f}(x,z)$ instead. Then we can again substitute $Y_f(x,z)$ with the corresponding $B_{n,k}^f$. We observe:
\begin{align*}
[z^n]f(x+z)^k&=[z^n]\left(f(x+z)-f(x)+f(x)\right)^k\\
&=[z^n]\sum_{j=0}^{k}\binom{k}{j}\left(f(x+z)-f(x)\right)^j\left(f(x)\right)^{k-j}\\
&=\left(f(x)\right)^k\delta_{n,0}+\sum_{j=1}^{k}\binom{k}{j}\left(f(x)\right)^{k-j}[z^n]\left(f(x+z)-f(x)\right)^j\\
&=\left(f(x)\right)^k\delta_{n,0}+\sum_{j=1}^{k}\binom{k}{j}\left(f(x)\right)^{k-j}Y_f^\triangle(n,j,x)\tag{16}
\end{align*}
Since $Y_f^\triangle(n,j,x)$ is defined for $n\geq 1$ only we have to consider the case $j=0$ separately. We use thereby the Kronecker $\delta$-symbol. Now substituting (16) in (15) gives:
\begin{align*}
Y_{f\cdot h}^{\triangle}(n,k,x)&=\sum_{j=0}^{k}\binom{k}{j}\left(-f(x)h(x)\right)^{k-j}\sum_{l=0}^{n}\\
&\qquad\left(\left(f(x)\right)^j\delta_{l,0}+\sum_{m_1=0}^{j}\binom{j}{m_1}\left(f(x)\right)^{j-m_1}Y_f^\triangle(l,m_1,x)\right)\\
&\qquad\cdot\left(\left(h(x)\right)^j\delta_{n-l,0}+\sum_{m_2=0}^{j}\binom{j}{m_2}\left(h(x)\right)^{j-m_2}Y_h^\triangle(n-l,m_2,x)\right)\\
&=\left(f(x)\right)^k\sum_{j=0}^{k}\binom{k}{j}\left(-1\right)^{k-j}\sum_{m_2=1}^{j}\binom{j}{m_2}\left(h(x)\right)^{k-m_2}Y_h^\triangle(n,m_2,x)\\
&\qquad+\sum_{j=0}^{k}\binom{k}{j}(-1)^{k-j}\sum_{l=1}^{n-1}
\left(\sum_{m_1=0}^{j}\binom{j}{m_1}\left(f(x)\right)^{k-m_1}Y_f^\triangle(l,m_1,x)\right)\\
&\qquad\qquad\cdot\left(\sum_{m_2=0}^{j}\binom{j}{m_2}\left(h(x)\right)^{k-m_2}Y_h^\triangle(n-l,m_2,x)\right)\\
&\qquad+\left(h(x)\right)^k\sum_{j=0}^{k}\binom{k}{j}\left(-1\right)^{k-j}\sum_{m_1=1}^{j}\binom{j}{m_1}\left(f(x)\right)^{k-m_1}Y_f^\triangle(n,m_1,x)\\
\end{align*}
Substituting $B_{n,k}^{f\cdot h}(x)$ for $Y_{f\cdot h}^{\triangle}(n,k,x)$ (and omitting the argument $x$) gives:
\begin{align*}
k!\frac{B_{n,k}^{f\cdot h}}{(fh)^{k}}&=\sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j}
\left(\sum_{m=1}^j\binom{j}{m}m!\left(\frac{B_{n,m}^{h}}{h^{m}} + \frac{B_{n,m}^{f}}{f^{m}}\right)\right.\\
&\qquad+\sum_{l=1}^{n-1}\binom{n}{l}\left(\sum_{m_1=1}^j\binom{j}{m_1}m_{1}!\frac{B_{l,m_1}^f}{f^{m_1}}\right)
\left.\left(\sum_{m_2=1}^j\binom{j}{m_2}m_{2}!\frac{B_{n-l,m_2}^h}{h^{m_2}}\right)\right)\tag{17}\\
\end{align*}
with the factor $\frac{k!}{(fh)^k}$ on the left-hand side to better show the nice symmetry.
Step 2.3.2: Simplification of $B^{f\cdot h}_{n,k}$
We can simplify (17) considerably. Observe, that the RHS consists of two parts, the first being a double sum with structure
\begin{align*}
\sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j}\sum_{m=1}^j\binom{j}{m}\varphi_m \qquad\qquad k\geq 1
\end{align*}
and $\varphi_m$ equal to
\begin{align*}
\varphi_m=m!\left(\frac{B_{n,m}^{f}}{f^{m}} + \frac{B_{n,m}^{h}}{h^{m}}\right)\tag{18}
\end{align*}
We show that following is valid
\begin{align*}
\sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j}\sum_{m=1}^j\binom{j}{m}\varphi_m=\varphi_k \qquad\qquad k\geq 1\tag{19}
\end{align*}
We observe
\begin{align*}
\sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m=1}^j\binom{j}{m}\varphi_m\\
&=\sum_{m=1}^k\varphi_m\sum_{j=m}^{k}\binom{k}{j}\binom{j}{m}(-1)^{k-j}\\
&=\sum_{m=1}^k\varphi_m\sum_{j=m}^{k}\binom{k}{m}\binom{k-m}{j-m}(-1)^{k-j}\\
&=\sum_{m=1}^k\binom{k}{m}\varphi_m\sum_{j=0}^{k-m}\binom{k-m}{j}(-1)^{(k-m)-j}\\
&=\sum_{m=1}^k\binom{k}{m}\varphi_m\delta_{{k-m},0}\\
&=\varphi_k
\end{align*}
and (19) follows.
Now we take a look at the second part of the RHS of (17). We identify following structure
\begin{align*}
\sum_{j=1}^{k}\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2} \qquad\qquad k\geq 1
\end{align*}
with $\psi_{m_1},\psi_{m_2}$ equal to
\begin{align*}
\psi_{m_1}=m_{1}!\frac{B_{i,m_1}^f}{f^{m_1}}\qquad\qquad\psi_{m_2}=m_{2}!\frac{B_{n-i,m_2}^{h}}{h^{m_2}}\tag{20}
\end{align*}
The sum $\sum_{i=1}^{n-1}\binom{n}{i}$ is not part of our consideration. It does neither depend on $\psi_{m_1}$ nor on $\psi_{m_2}$ and we think of it as being at the front of the RHS (outside from $\sum_{j=1}^{k}$).
We show the following is valid
\begin{align*}
\sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\\
&\quad=\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}\psi_{m_1}\psi_{m_2}\qquad\qquad k\geq 1\tag{21}
\end{align*}
This identity is less simple than the identity (19) above but with the proper tool at hand we can prove this identity en passant. We will use Egorychevs residual calculus of formal power series.
Note: This powerful technique is based upon Cauchys residue theorem and was introduced by G.P. Egorychev (Integral Representation and the Computation of Combinatorial Sums) to compute binomial identies.
We use only two aspects of this theory:
Let $A(z)=\sum_{j=0}^{\infty}a_jz^j$ be a formal power series, then
- Write the binomial coeffients as residuals of corresponding formal power series
\begin{align*}
\mathop{res}_{z}\frac{A(z)}{z^{j+1}}=a_j\tag{22}
\end{align*}
- Apply the substitution rule for formal power series:
\begin{align*}
A(z)=\sum_{j=0}^{\infty}a_jz^{j}=\sum_{j=0}^{\infty}z^j\mathop{res}_{w}\frac{A(w)}{w^{j+1}}\tag{23}
\end{align*}
We start the proof with elementary transformations:
\begin{align*}
\sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\\
&=\sum_{m_1=1}^{k}\psi_{m_1}\sum_{j=m_1}^k\binom{k}{j}\binom{j}{m_1}(-1)^{k-j}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\tag{24}\\
&=\sum_{m_1=1}^{k}\psi_{m_1}\left(\sum_{m_2=1}^{m_1}\psi_{m_2}\sum_{j=m_1}^k\binom{k}{j}\binom{j}{m_1}\binom{j}{m_2}(-1)^{k-j}\right.\\
&\qquad\qquad\left.+\sum_{m_2=m_1+1}^{k}\psi_{m_2}\sum_{j=m_2}^k\binom{k}{j}\binom{j}{m_1}\binom{j}{m_2}(-1)^{k-j}\right)\tag{25}\\
&=\sum_{m_1=1}^{k}\psi_{m_1}\left(\sum_{m_2=1}^{m_1}\psi_{m_2}\sum_{j=m_1}^k\binom{k}{m_1}\binom{k-m_1}{j-m_1}\binom{j}{m_2}(-1)^{k-j}\right.\\
&\qquad\qquad\left.+\sum_{m_2=m_1+1}^{k}\psi_{m_2}\sum_{j=m_2}^k\binom{k}{m_2}\binom{k-m_2}{j-m_2}\binom{j}{m_1}(-1)^{k-j}\right)\tag{26}\\
&=\sum_{m_1=1}^{k}\psi_{m_1}\left(\binom{k}{m_1}\sum_{m_2=1}^{m_1}\psi_{m_2}\sum_{j=0}^{k-m_1}\binom{k-m_1}{j}\binom{j+m_1}{m_2}(-1)^{(k-m_1)-j}\right.\\
&\qquad\qquad\left.+\sum_{m_2=m_1+1}^{k}\binom{k}{m_2}\psi_{m_2}\sum_{j=0}^{k-m_2}\binom{k-m_2}{j}\binom{j+m_2}{m_1}(-1)^{(k-m_2)-j}\right)\tag{27}\\
\end{align*}
Comment:
(24) Exchange of the sums with indices $j$ and $m_1$.
(25) Exchange of the sums with indices $j$ and $m_2$.
(26) Usage of the binomial identity $\binom{k}{j}\binom{j}{m}=\binom{k}{m}\binom{k-m}{j-m}$
(27) Index shift $j\rightarrow j-m_1,j\rightarrow j-m_2$
We now apply Egorychevs technique to show that
The following is valid (and symmetrically with $m_1,m_2$ exchanged):
\begin{align*}
\sum_{j=0}^{k-m_1}\binom{k-m_1}{j}\binom{j+m_1}{m_2}(-1)^{(k-m_1)-j}=\binom{m_1}{k-m_2}\qquad\qquad k\geq 1\tag{28}
\end{align*}
We observe
\begin{align*}
\sum_{j=0}^{k-m_1}&\binom{k-m_1}{j}\binom{j+m_1}{m_2}(-1)^{(k-m_1)-j}\\
&=\sum_{j\geq 0}\binom{k-m_1}{j}\binom{-(m_2+1)}{j+m_1-m_2}(-1)^{k-m_2}\tag{29}\\
&=(-1)^{k-m_2}\sum_{j\geq 0}\mathop{res}_{z}\frac{(1+z)^{k-m_1}}{z^{j+1}}
\mathop{res}_u\frac{(1+u)^{-(m_2+1)}}{u^{j+m_1-m_2+1}}\tag{30}\\
&=(-1)^{k-m_2}\mathop{res}_u\frac{(1+u)^{-(m_2+1)}}{u^{m_1-m_2+1}}\sum_{j\geq 0}u^{-j}\mathop{res}_{z}\frac{(1+z)^{k-m_1}}{z^{j+1}}\tag{31}\\
&=(-1)^{k-m_2}\mathop{res}_u\frac{(1+u)^{-(m_2+1)}}{u^{m_1-m_2+1}}\left(1+\frac{1}{u}\right)^{k-m_1}\tag{32}\\
&=(-1)^{k-m_2}\mathop{res}_u\frac{(1+u)^{k-m_2-m_1-1}}{u^{k-m_2+1}}\tag{33}\\
&=(-1)^{k-m_2}\binom{k-m_2-m_1-1}{k-m_2}\tag{34}\\
&=\binom{m_1}{k-m_2}\tag{35}
\end{align*}
Comment:
In (28) we use $\binom{-n}{k}=\binom{n+k-1}{k}(-1)^k$ and changed the limit to $\infty$ without changing the value, since we add only $0$.
In (29) we use $(1+w)^n$ as formal power series with the coefficients $\binom{n}{k}$ according to (22). Observe that $\binom{k-m_2}{j}$ is the coefficient of $[z^{j}]$ in $(1+z)^{k-m_1}$ and similarly for $u$.
In (30) we do some rearrangement to prepare the application of the substitution rule
In (31) we apply the substitution rule according to (23)
In (32) there is a simple rearrangement
In (33) we do the same as we did from (29) to (30) but in reverse direction.
We now proceed substituting the identity (28) in (27) and get:
\begin{align*}
\sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}\psi_{m_1}\sum_{m_2=1}^j\binom{j}{m_2}\psi_{m_2}\\
&=\sum_{m_1=1}^{k}\left(\sum_{m_2=1}^{m_1}\binom{k}{m_1}\binom{m_1}{k-m_2}\psi_{m_1}\psi_{m_2}\right.\\
&\qquad\left.+\sum_{m_2=m_1+1}^k\binom{k}{m_2}\binom{m_2}{k-m_1}\psi_{m_1}\psi_{m_2}\right)\\
&=\sum_{m_1=1}^{k}\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}\psi_{m_1}\psi_{m_2}\\
\end{align*}
In the last line we use the identity $\binom{k}{m_1}\binom{m_1}{k-m_2}=\binom{k}{m_2}\binom{m_2}{k-m_1}$
and (21) follows.
Let's summarise:
According to (18) and (19) we get
\begin{align*}
\sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m=1}^{j}\binom{j}{m}
m!\left(\frac{B_{n,m}^{f}}{f^{m}} + \frac{B_{n,m}^{h}}{h^{m}}\right)\\
&=k!\left(\frac{B_{n,k}^{f}}{f^{k}} + \frac{B_{n,k}^{h}}{h^{k}}\right)\qquad\qquad k \geq 1\tag{34}
\end{align*}
and according to (20) and (21) we get
\begin{align*}
\sum_{j=1}^{k}&\binom{k}{j}(-1)^{k-j}\sum_{m_1=1}^j\binom{j}{m_1}m_{1}!\frac{B_{i,m_1}^f}{f^{m_1}}\sum_{m_2=1}^j\binom{j}{m_2}m_{2}!\frac{B_{n-i,m_2}^{h}}{h^{m_2}}\\
&\quad=\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}!m_{2}!\frac{B_{i,m_1}^f}{f^{m_1}}\frac{B_{n-i,m_2}^{h}}{h^{m_2}}\qquad\qquad k\geq 1\tag{35}\\
\end{align*}
With the help of (34) and (35) the identity (17) simplifies to
\begin{align*}
k!\frac{B_{n,k}^{f\cdot h}}{(fh)^{k}}&=k!\left(\frac{B_{n,k}^{f}}{f^{k}} + \frac{B_{n,k}^{h}}{h^{k}}\right)+\sum_{l=1}^{n-1}\binom{n}{l}\\
&\qquad\cdot\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}!m_{2}!\frac{B_{l,m_1}^f}{f^{m_1}}\frac{B_{n-l,m_2}^{h}}{h^{m_2}}\\
\end{align*}
We finally get
\begin{align*}
B_{n,k}^{f\cdot h}&=h^k B_{n,k}^{f} + f^k B_{n,k}^{h}\\
&\qquad+\frac{1}{k!}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}! m_{2}!f^{k-m_1}h^{k-m_2}\\
&\qquad\qquad\cdot \sum_{l=1}^{n-1}\binom{n}{l}B_{l,m_1}^fB_{n-l,m_2}^{h}\tag{36}\\
\end{align*}
Step 2.4: Calculation of $B^{f\cdot (\ln \circ g)}_{n,k}$
Now it's time to harvest the first time. We can now combine the expression with Bell polynomials of $f\cdot h$ and of $(\ln\circ g)$ to
calculate the Bell polynomials of $f\cdot (\ln\circ g)$.
We conclude:
\begin{align*}
B_{n,k}^{f\cdot (\ln \circ g)}&= (\ln \circ g)^k B_{n,k}^{f} + f^k B_{n,k}^{\ln \circ g}\\
&\qquad+\frac{1}{k!}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\binom{k}{m_1}\binom{m_1}{k-m_2}m_{1}! m_{2}!f^{k-m_1} (\ln \circ g)^{k-m_2}\\
&\qquad\qquad\cdot\sum_{l=1}^{n-1}\binom{n}{l}B_{l,m_1}^fB_{n-l,m_2}^{\ln \circ g}\tag{37}
\end{align*}
According to (14) we know
\begin{align*}
B^{\ln\circ g}_{n,k}&=\sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}
\end{align*}
Substituting this expression in (37) and summing over $k=1,\ldots,n$ gives:
\begin{align*}
\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)}
&=\sum_{k=1}^{n}\left((\ln \circ g)^k B_{n,k}^{f} + f^k \sum_{j=k}^{n}\frac{(-1)^{j-k}}{g^j}\begin{bmatrix}j\\k\end{bmatrix}B^{g}_{n,j}\right)\\
&\qquad+\sum_{k=1}^{n}\sum_{m_1=1}^k\sum_{m_2=1}^{k}\frac{m_2!}{(k-m_1)!}\binom{m_1}{k-m_2}f^{k-m_1} (\ln \circ g)^{k-m_2}\\
&\qquad\qquad\cdot\sum_{l=1}^{n-1}\sum_{j=m_2}^{n-l}\frac{(-1)^{j-m_2}}{g^j}\begin{bmatrix}j\\m_2\end{bmatrix}\binom{n}{l}B_{l,m_1}^fB^{g}_{n-l,j}\tag{38}
\end{align*}
Summary of part 2: With expression (38) we have developed a formula for $\sum_{k=1}^{n}B_{n,k}^{f\cdot (\ln \circ g)}$ and part 2 is finished. The last challenge is to show that OPs expression is equal to this one. This is the theme of part 3.
Note: The body of this answer is limited with $30000$ chars. I'll provide part 3 in a follow-up answer.
