In these situations, the Mean-value Theorem/inequality (whatever you want to call it) is your best friend. What I'm about to write, I learnt from Henri Cartan's book on Differential Calculus. The relevant results are Theorem $3.11$ (Mean value theorem) and Corollary $5.53$, and Theorem $5.62$ (Taylor's theorem with Lagrange Remainder).
Theorem $3.11$ (Mean-Value Theorem):
Let $E$ be any real Banach space, $[a,b]$ an interval in $\Bbb{R}$ with $a< b$. Let $u:[a,b] \to E$ and $g:[a,b] \to \Bbb{R}$ be continuous functions on $[a,b]$, and differentiable on the open interval $(a,b)$. Suppose that for all $t \in (a,b)$, it is true that
\begin{align}
\lVert u'(t)\rVert & \leq g'(t)
\end{align}
Then, $\lVert u(b) - u(a)\rVert \leq g(b) - g(a)$.
The proof of this theorem, while elementary in methods, is slightly technical. But, you said you know how to deduce the second inequality from the first, so I guess you must know of the Mean value theorem either in this form or something similar, so I'll omit the proof.
Next, we have the following result (Proposition $5.51$ and Corollary $5.53$ combined)
Theorem:
Let $E$ be a real Banach space, $I$ an open subset of $\Bbb{R}$ containing the closed interval $[0,1]$, and let $v: I \to E$ be $(n+1)$-times differentiable on $I$, and suppose that $M:= \sup\limits_{t \in [0,1]} \lVert v^{(n+1)}(t)\rVert < \infty$. (Note that boundedness of derivative is obviously a weaker assumption than the $(n+1)^{th}$ derivative being continuous on $I$). Then the $n^{th}$ order remainder of $v$ at $0$ satisfies:
\begin{align}
\left \lVert v(t) - \sum_{k=0}^n \dfrac{v^{(k)}(0)}{k!} t^k\right \rVert &\leq \dfrac{M}{(n+1)!}. \tag{$\ddot{\smile}$}
\end{align}
This will be the crucial link to help you deduce the theorem you want, because, we'll apply it to $v(t) = f(x + th)$, for $t$ in an open interval containing $[0,1]$. Here's the proof of this theorem:
Define the functions $u: I \to E$, and $g: I \to \Bbb{R}$ by
\begin{align}
u(t) &= \sum_{k=0}^n \dfrac{(1-t)^k}{k!} v^{(k)}(t) \quad \text{and} \quad
g(t) = -M \dfrac{(1-t)^{n+1}}{(n+1)!}
\end{align}
Then a simple calculation using the product rule (there will be a telescoping sum) shows that
\begin{align}
u'(t) &= \dfrac{(1-t)^n}{n!} v^{(n+1)}(t)
\end{align}
Hence, for all $t \in [0,1]$, we have
\begin{align}
\lVert u'(t)\rVert & \leq M \dfrac{(1-t)^n}{n!} = g'(t)
\end{align}
This is exactly the situation of the Mean-value theorem, so, we have $\lVert u(1) - u(0)\rVert \leq g(1) - g(0)$. If you plug in what $u$ and $g$ are, you'll find exactly $(\ddot{\smile})$. (The $u$ was constructed to ensure exactly this).
Lastly, we let $v(t) = f(x+th)$, where the $f,x,h$ are as in your question. Then, the $(n+1)^{th}$ derivative of $v$ at $t$ is given by (chain rule, and induction)
\begin{align}
v^{n+1}(t) &= (D^{n+1}f)_{x+th} [h]^{n+1} \in E
\end{align}
Btw here, the RHS means the $(n+1)^{th}$ Frechet derivative of $f$ at $x+th$ evaluated on the $n+1$ tuple $(h, \dots, h) \in E^{n+1}$. Now, we can bound this using the operator norm, and properties of continuous multi-linear maps:
\begin{align}
\lVert v^{n+1}(t)\rVert & \leq \lVert D^{n+1}f{(x+th)}\rVert \cdot \lVert h\rVert^{n+1} \\
& \leq M \cdot \lVert h \rVert^{n+1}
\end{align}
So, if in $(\ddot{\smile})$ you now substitute what $v$ is, and replace $M$ by $M \cdot \lVert h\rVert^{n+1}$, then you find that
\begin{align}
\left \lVert f(x+th) - \sum_{k=0}^n \dfrac{D^kf_x(th)^k}{k!} \right \rVert & \leq \dfrac{M}{(n+1)!}\lVert h \rVert^{n+1}
\end{align}
which is exactly the bound on the remainder term.
