Let $E$ be a normed $\mathbb R$-vector space, $\Omega\subseteq E$ be open, $x_0\in\Omega$, $k\in\mathbb N$ and $f:\Omega\to\mathbb R$ be $(k-1)$-times Fréchet differentiable and $k$-times Fréchet differentiable at $x_0$.
If $E=\mathbb R$, we easily see that $$\forall x\in\Omega:f(x)=\sum_{i=0}^kf^{(i)}(x_0)\frac{(x-x_0)^i}{i!}+h(x)(x-x_0)^k\tag1$$ for some $h:\Omega\to\mathbb R$ which is continuous at $x_0$. In fact, we can simply choose $$h(x):=\begin{cases}\displaystyle\frac{f(x)-\sum_{i=0}^kf^{(i)}(x_0)\frac{(x-x_0)^i}{i!}}{(x-x_0)^k}&\text{, if }x\in\Omega\setminus\{x_0\}\\0&\text{, if }x=x_0\end{cases}$$ for $x\in\Omega$.
Can we generalize this? At least when $E$ is a $\mathbb R$-Hilbert space?
The proof for the case $E=\mathbb R$ and $k\ge2$ requires L'Hôpital's rule, which is why I don't know how to generalize this result (unless $k=1$, where we can choose $h$ similar to the case $E=\mathbb R$, but with $x-x_0$ replaced by $\|x-x_0\|_E$ replaced in the denominator in the definiton of $h$ and in the factor after $h(x)$ in the formula for $f(x)$.)
EDIT:
We've got the following mean-value theorem: Let $F$ be another normed $\mathbb R$-vector space and $g:\Omega\to F$. If $x,y\in\Omega$ with $[x,y]:=\{(1-t)x+ty:t\in[0,1]\}\subseteq\Omega$ and $g$ is continuous on $[x,y]$ and differentiable in $(x,y):=\{(1-t)x+ty:t\in(0,1)\}$, then $$\exists z\in(x,y):\|g(x)-g(y)\|_F\le\|{\rm D}g(z)(x-y)\|_F.\tag2$$ If $F=\mathbb R$, then $$\exists z\in(x,y):g(x)-g(y)={\rm D}g(z)(x-y)\tag{2'}.$$ Can we use this to generalize the result?
EDIT 2:
Let's try to prove this by induction: If $k=1$, then we can choose $$h(x):=\begin{cases}\displaystyle\frac{f(x)-f(x_0)-{\rm D}f(x_0)(x-x_0)}{\|x-x_0\|_E}&\text{, if }x\in\Omega\setminus\{x_0\}\\0&\text{, if }x=x_0\end{cases}$$ for $x\in\Omega$ and obtain $$\forall x\in\Omega:f(x)=f(x_0)+{\rm D}f(x_0)(x-x_0)+h(x)\|x-x_0\|_E\tag3.$$ $h$ is continuous at $x_0$ by definition of Fréchet differentiability at $x_0$.
Now, for simplicity, consider $k=2$. Following the approach described by peek-a-boo in the comments, let $$T(x):=f(x_0)-{\rm D}f(x_0)(x-x_0)-\frac12{\rm D}^2f(x_0)(x-x_0)^2\;\;\;\text{for }x\in E$$ and $$\varphi(x):=f(x)-T(x)\;\;\;\text{for }x\in\Omega.$$ Since $f$ is differentiable, $\varphi$ is differentiable as well and $${\rm D}\varphi(x)={\rm D}f(x)-{\rm D}f(x_0)-{\rm D}^2f(x_0)(x-x_0)\tag4.$$ Since $f$ is twice differentiable at $x_0$, $$\frac{\left\|{\rm D}\varphi(x)\right\|_{E'}}{\|x-x_0\|_E}\xrightarrow{x\to x_0}0\tag5.$$ Since $\Omega$ is open, there is a $\varepsilon>0$ with $B_\varepsilon(x_0)\subseteq\Omega$. Let $(x_n)_{n\in\mathbb N}\subseteq B_\varepsilon(x_0)\setminus\{x_0\}$ with $x_n\xrightarrow{n\to\infty}x_0$. By the mean value theorem, there is a $(\xi_n)_{n\in\mathbb N}\subseteq\Omega$ with $$\xi_n\in(x_n,x_0)\tag6$$ and $$|\varphi(x_n)-\varphi(x_0)|\le\|{\rm D}\varphi(\xi_n)\|_{E'}\|x_n-x_0\|_E\tag7$$ for all $n\in\mathbb N$. Now, $$\frac{|f(x_n)-T(x_n)|}{\|x_n-x_0\|_E^2}=\frac{|\varphi(x_n)-\varphi(x_0)|}{\|x_n-x_0\|_E^2}\le\frac{\|{\rm D}\varphi(\xi_n)\|_{E'}}{\|x_n-x_0\|_E}\tag8$$ for all $n\in\mathbb N$.
But does the right-hand side of $(8)$ tend to $0$? Clearly, $\xi_n\xrightarrow{n\to\infty}x_0$ and ${\rm D}\varphi$ is continuous at $x_0$. However, the right-hand side of $(8)$ doesn't match $(5)$ ...