As I said in the comments, you can do this by pure algebra, but it's not
entirely straightforward and a bit yucky. I think your sum should be
$-1 + \sum_{k = 1}^n \operatorname{IC}(n, k) \cdot (k + 1)$,
because for each $k$, for each composition of $n$ with $k$ parts, you have
$k + 1$ positions to insert the $n$-part, and this double-counts $n + n$. This
sum is equal to $-1 + \sum_{k = 1}^n \binom{n - 1}{k - 1}(k + 1)$, and you can
evaluate this by using standard tricks/identities for sums of binomial
coefficients. I will go into more detail than that for now - at the end of this answer.
Here is what I think is quite a nice combinatorial "bijective" viewpoint:
In general a nice way to get a combinatorial foothold with compositions is to
use the fact that the compositions of $n$ correspond to the subsets
of $\{1, \dotsc, n - 1\}$, by identifying a composition
$n_1 + \dotsb + n_k$ with the set $X$ of its partial sums, excluding $0$ and
$n$: $X = \{n_1, n_1 + n_2, \dotsc, n_1 + \dotsb + n_{k - 1}\}$.
This identification is not new - it's used, for instance in the nicer bijective
way to prove that the number of compositions of $n$ is $2^{n - 1}$. It's a
slightly more formal formulation of a "stars-and-bars" argument. This has been
discussed before on this site, for example
here. I have a mild personal
preference for using concrete sets of naturals for proofs like this because I
feel like it helps me avoid off-by-one errors. If you prefer the more typical
combinatorial type of proof, you can easily translate the following discussion
into triangles and squares, or stars and bars, or goats and pigs, or whatever
type of objects you like! Anyway, on with the answer...
We'll count separately "the compositions of $2n$ with first or last part
$n$" and "the compositions of $2n$ with largest part $n$, but it's not the first
or last part".
Under the above identification, compositions of $2n$ with maximum part $n$,
which is not the first or last part, correspond to subsets $X$ of
$\{1, \dotsc, 2n - 1\}$, where there is some $1 \le k \le n - 1$ such that
$k, k + n \in X$ and $k + t \notin X$ for all $1 \le t \le n - 1$. The value of
$k$ for such a set $X$ is unique (because there is only one part of size $n$).
So we can count the number of such $X$ by first choosing a $k$ and then choosing
all suitable $X$ for that $k$.
There are $n - 1$ possible choices for $k$, and given $k$, there are
$2^{2n - 1 - (n + 1)} = 2^{n - 2}$ choices for $X$
(since we've already determined whether $k, \dotsc, k + n$ belong to $X$,
and we may freely choose for the remainder of the elements).
So there are $(n - 1)2^{n - 2}$ such compositions.
Compositions of $2n$ with first part $n$ of course correspond to compositions of
$n$. (As do compositions with last part $n$). This double-counts the composition
$n + n$, so the total number of remaining compositions is
$2^{n - 1} + 2^{n - 1} - 1$.
Hence, in total, the number of such compositions is
$(n - 1)2^{n - 2} + 2^n - 1 = (n + 3)2^{n - 2} - 1$.
More on the summation approach. I'm taking it for granted that you know that $\mathrm{IC}(n, k) = \binom{n - 1}{k - 1}$. (In fact this follows from the bijective viewpoint - a $k$-composition corresponds to a subset of size $k - 1$.)
We'll also need the fact that $\sum_{r = 0}^m r\binom mr = m \cdot 2^{m - 1}$. There's a cute proof given here (essentially by integrating the generating function). There is also a nice bijective way to see it here. In that thread there are many other approaches too, including this nice elementary approach.
This lets us work out the sum. First change variables to $r = k - 1$:
\begin{align*}
\sum_{k = 1}^n \binom{n - 1}{k - 1}(k + 1)
&= \sum_{r = 0}^{n - 1} \binom{n - 1}r(r + 2) \\
&= \sum_{r = 0}^{n - 1} r\binom{n - 1}r + 2\sum_{r = 0}^{n - 1} \binom{n - 1}r \\
&= (n - 1)2^{n - 2} + 2 \cdot 2^{n - 1}
\end{align*}
which gives the correct result after subtracting the $1$.
In this case if you wanted to prove it totally manually, the first approach would lead you to consider $f(x) = \sum_{k = 1}^n \binom{n - 1}{k - 1}(k + 1)x^k$, then find an antiderivative $F(x)$ of $f(x)$, write $F(x)$ in closed form using the binomial theorem, and differentiate to find $f(1)$.
