Does differentiability have a geometric interpretation for high dimensional functions? A function $f:\mathbb{R^n} \to \mathbb{R^m}$ is said to be differentiable if the limit $$ \lim_{x \to a} \dfrac{||f(x) - f(a) - D_{f(a)}(x-a) ||}{||x-a ||}=0$$
exists where $D_{f(a)}$ is the $m \times n$ matrix of partial derivatives of the function $f$. Now, if the function is a scalar valued function, where $m=1$ then there is a clear geometric interpretation: $L(x)= f(a) + D_{f(a)}(x-a)$ is the function of the linear approximation for $f$ in the neighborhood of $a$. It is a line for $n=1$, a plane for $n=2$ and hyperplane for higher dimensional input spaces.
Is there such an interpretation as well when $f$ is a vector function where $m>1$? When this is the case and $f(x)=(u_1(x),\dots,u_m(x))^T$, then the definition of the derivative defines separate linear approximations for each $u_i(x)$ component of $f(x)$ via the each row of the partial derivative matrix $D_{f(a)}$. This looks like a straightforward expansion of the scalar case to vector valued functions. But I cannot imagine a geometric justification of this like in the scalar case (where $m=1$). So, is this just an obvious generalization of the definition of differentiability to vector valued functions or does it have a geometric justification as well, like in the $m=1$ case? (Linear approximation around $a$)
 A: The "implicit" way of defining $df(a)$ by the formula in your question is exactly equivalent to the idea that $df(a)$ should provide a linear approximation to the true increment of $f$ when moving away from $a$ by a small (vectorial) amount $X$. For any $X$, small or large, one has
$$f(a+X)-f(a)=df(a).X + R(X)\ ,\tag{1}$$
where you may regard this equation as definition of the error $R(X)$.The essential point now is the following: "Generically" the term $df(a).X$, being linear in $X$, will have the order of magnitude $X$. Therefore it only makes sense to consider $df(a).X$ as a linear approximation to the true increment, resp., to neglect the error term $R(X)$ in $(1)$, if we can be sure that $|R(X)|$ is way smaller than this  linear term $df(a).X\ $. 
Therefore the proper definition of differentiability reads as follows: The function $f$ is differentiable at $a$ if there exists a linear map $L=:df(a)$ such that
$$f(a+X)-f(a)=L.X+ o(|X|)\qquad(X\to0)\ .\tag{2}$$
(Here $\ $ "$+o(|X|)$" $\ $ is shorthand for$\ $ "$+R(X)$, whereby $\lim_{X\to0}{R(X)\over|X|}=0$".) From $(2)$ it follows that
$$\lim_{X\to0}{f(a+X)-f(a)-df(a).X\over |X|}=\lim_{X\to0}{R(X)\over|X|}=0\ ,$$ as in your definition. We don't have to talk about coordinates when dealing with such facts, because we all know that $X\to0$ is equivalent with $X_k\to0$ for each $k\in[n]$ individually, same for $f=(f_1,\ldots, f_n)$.
