Clarification on definition of differentiability of vector-valued functions. As it is normally stated, the definition of differentiability of vector-valued function is as follows:
$\textbf{Definition:}$ Let $U \subset \Bbb{R}^n$ be open, and let $a \in U$. A vector-valued function $f$ is differentiable at $a$ if there is a linear map $Df(a): \Bbb{R}^n \to \Bbb{R}^n$ so that
$$\lim\limits_{h \to 0} \frac{f(a+h)-f(a)-Df(a)h}{\Vert h \Vert} = 0.$$
My question is, is the vector $h$ implicitly assumed to be arbitrary such that it is analogous for the case of real-valued functions where the direction of approaching the limit point should not matter?

Yes, $h$ is whatever vector of any direction in the definition of differentiability. You just have obviously to have $a+h \in U$.
