In the Wikipedia entry for "Differentiable Function" there is the following section on "Differentiability in Higher Dimensions":
A function of several variables $\mathbb{R}^m\to\mathbb{R}^n$ is said to be differentiable at a point $\pmb{x_0}$ if there exists a linear map $\pmb{J:\mathbb{R}^m\to\mathbb{R}^n}$ such that
$$\lim\limits_{\pmb{h}\to \pmb{0}} \frac{\Vert \pmb{f}(\pmb{x_0}+\pmb{h})-\pmb{f}(\pmb{x_0})-\pmb{J(h)}\rVert_{\mathbb{R}^n}}{\lVert \pmb{h} \rVert_{\mathbb{R}^m}}=0$$
If a function is differentiable at $\pmb{x_0}$, then all of the partial derivatives exist at $\pmb{x_0}$, and the linear map $\pmb{J}$ is given by the Jacobian matrix, an $n x m$ matrix in this case.
My question is about the meaning of partial derivative and also continuous differentiability as these concepts relate to a vector field.
If we have a scalar field $g:\mathbb{R}^m\to\mathbb{R}$, then I understand the partial derivatives to be the derivatives of $g$ relative to the standard basis vectors of the coordinate system being used. In this example, there are $m$ of them, and they are each scalar fields from $\mathbb{R}^m$ to $\mathbb{R}$.
In the case of $f:\mathbb{R}^m\to\mathbb{R}^n$, $f$ is a vector with $n$ components, each of which is a scalar field.
My first question is: what are the partial derivatives of $f$ at some $\pmb{x_0}$?
My guess is it is an $n x m$ matrix, where each row contains the partial derivatives of each component of $f$. Is this true?
My second question is about continuous differentiability.
In the case of a scalar field like $g$, we say it is continuously differentiable at a point $\pmb{x_0}$ if there is some n-ball centered at $\pmb{x_0}$ in which the partial derivatives are defined and, in addition, these partial derivatives are continuous at $\pmb{x_0}$.
My second question is: What is the exact definition of continuous differentiability of $f$, a vector field?
My guess is that each component of $f$ must be continuously differentiable at $\pmb{x_0}$. Is this true?
If $f\colon U \to \mathbb R^m$ is a function defined on an open subset of $\mathbb R^n$, then $f$ is differentiable at $a \in U$ if there is a linear map $\alpha \in \mathrm{Hom}(\mathbb R^n,\mathbb R^m)$ such that $$ f(a+h) = f(a)+\alpha(h) + o(\|h\|), $$ where the notation $o(\|h\|)$ denotes a function which lies in the subspace of functions $g \colon U \to \mathbb R^m$ such that $\|g(a+h)\|/\|h\|\to 0$ as $h \to 0$. If such an $\alpha$ exists it is unique and denoted $Df_a$ (or $Df(a)$ or....).
This is just a rephrasing of the definition given in the question, but it perhaps help emphasize that the definition requires $\alpha$, the derivative of $f$ must be the best linear approximation to $f(a+h)-f(a)$. To see why that is the case, note that any nonzero linear map $\beta\colon \mathbb R^n \to \mathbb R^m$ has operator norm
$$ \|\beta\|_{\infty} = \sup\{\|\beta(v)\|: v \in \mathbb R^n: \|v\|\leq 1\}>0, $$ and since the unit ball $B_n = \{x\in \mathbb R^n: \|x\|\leq 1\}$ is compact there is a vector $v_0$ with $\|\beta(v_0)\|=\|\beta\|_\infty$, and hence if we let $h = tv_0$ then $h \to 0$ as $t\to 0$, but $\|\beta(h)\|/\|h\| = \|\beta(v_0)\| = \|\beta\|_\infty$, so that $\beta(h)$ does not lie in $o(\|h\|)$. It follows that if $\alpha_1\colon \mathbb R^n \to \mathbb R^m$ is any linear map, $$ \begin{split} f(a+h)-\left(f(a) +\alpha_1(h)\right) &= \left( f(a+h)-f(a)-\alpha(h) \right) + (\alpha_1-\alpha)(h) \\ &= (\alpha_1-\alpha)(h) + o(\|h\|), \end{split} $$ so that the size of the error $\|h\|^{-1}.\|f(a+h)-f(a)-\alpha_1(h)\|$ in approximating $f(a+h)$ by $f(a)+\alpha_1(h)$ is $o(\|h|)$ only if $\alpha_1=\alpha$. If $\alpha_1 \neq \alpha$ then the relative error $\|f(a+h)-f(a)-\alpha_1(h)\|/\|h\|$ will be bounded but will fail to tend to zero as $h \to 0$.
Notice that if $f$ is differentiable at all $a \in U$, then we obtain a new function $Df\colon U \to \mathrm{Lin}(\mathbb R^n,\mathbb R^m)\cong \text{Mat}_{n,m} (\mathbb R)$. Since $\text{Mat}(\mathbb R^n,\mathbb R^m)$ is just a copy of $\mathbb R^{nm}$, we can therefor view $Df\colon U \to \mathbb R^{nm}$, whence it makes sense to ask if $Df$ is continuous or indeed even differentiable. We say that $f$ is $\mathcal C^1$ if $Df\colon U \to \mathrm{Mat}(\mathbb R^n,\mathbb R^m)$ exists for all $a \in U$ and is a continuous function. For example, if $f \colon \mathbb R^3\to \mathbb R^3$ is a vector field, then it is continuously differentiable if the partial derivatives $\partial_i f_j$ are continuous on $U$.
Finally, the most general definition of a partial derivative is exactly what you expect it to be: If $U$ is a subspace of $\mathbb R^n$, then you can ask whether $f$ has a best linear approximation at $a$ when you restrict $h$ to lie in the subspace $U$. If it does, that linear approximation is the partial derivative of $f$ in ``$U$-directions'', and it is often written as something like $\partial_U(f)(a)$.
Now if you have a basis of $\mathbb R^n$ (such as the standard basis $\{e_1,\ldots,e_n\}$, then this decomposes $\mathbb R^n$ into a direct sum of $n$ lines: $\mathbb R^n= \mathbb R.e_1\oplus \ldots \oplus \mathbb R.e_n$, and so we obtain $n$ partial derivatives, $\partial_i f(a) = \partial_{\mathbb R.e_i} f(a)$ (when they exist).
Now $\partial_i f(a)$ is a linear map from $\mathbb Re_i$ to $\mathbb R^m$, and since $\mathbb R.e_i$ is a one-dimensional vector space, that linear map is determined by the value it takes on the basis element $e_i$. That vector in $\mathbb R^m$ is, strictly speaking something that is called the directional derivative of $f$ at $a$ in the direction $e_i$, but by a pretty universal abuse of notation, it is also often identified with the partial derivative. The subtlety here is that it can be the case that all the partial derivatives exist at a point $a \in U$, yet $f$ is not differentiable there. The good news, however, is that if we require $f$ to be continuously differentiable, that is, $Df\colon U \to \mathrm{Lin}(\mathbb R^n, \mathbb R^m)$ exists and is continuous on $U$, then this is equivalent to all the partial derivatives being continuous.