I'm not sure if I posed the right question, but this is my curiosity:
That a function is differentiable in $P\in\mathbb{R}^n$ means that given $F:\mathbb{R}^n\rightarrow\mathbb{R}^m$
$$ \lim_{X\to P}\frac{\|F(X)-F(P)-J(X-P)\|}{\|X-P\|}=0 $$ I understand that in terms of functions that go from $\mathbb{R}^n\to\mathbb{R}$, that expression basically means that given a hyperplane that goes through the point $(P,f(P))$, as you approach that point along any curve of points $(X,f(X))$, the angle of the triangle with those two points as vertices and the coordinate of the plane in $X$ should approach $0$. That at least gives me an intuition of how the plane is actually approximating the function near that point. But,
$1)$ Why is that the definition of differentiation, exactly?
$2)$ Is there an intuition for the more general $\mathbb{R}^n\rightarrow\mathbb{R}^m$ type of function?
Concerning functions, that is $f:\mathbb R^n\to\mathbb R$, I'd rather put it this way. One understands derivation in one variable, hence starts by derivating the restrictions $f|\ell$ to lines $\ell$ through a point $P$. This is just a one variable limit. Then, reasonably enough one asks these limits exist for all lines $\ell$ defined by a non-zero vectors $u$ at the point (for the limit definition it is not necessary the vector to be unitary, and if $u=0$ the limit is trivially $0$). This are the directional derivatives $D_uf(P)$, which give the map $J(u)=D_uf(P)$. Geometrically, each directional derivative gives the (tangent) line that best approximate the section of the graph $\varGamma\subset\mathbb R^n\times\mathbb R$ of $f$ cut by the plane $\ell\times\mathbb R$. But nothing guarantees all these tangent lines form a hyperplane! Thus why one asks the function $J:\mathbb R^n\to\mathbb R$ (with $L(0)=0$) to be linear. But still one knows that in several variables a function can have a limit along all directions without having a global limit, hence one asks for the global limit condition $$ \lim_{u\to0}\frac{f(P+u)-f(P)-J(u)}{\|u\|}=0. $$ For a mapping $f:\mathbb R^n\to\mathbb R^m$ the variation is that the sections of the graph are by $(m+1)$-planes $\ell\times\mathbb R^m$ and you get more arbitrary curves (instead of planar). Still the derivative of that restriction gives the tangents to those curves. But this enters more the area of a curves and surfaces course. From the Analysis viewpoint I'd resource to the work componentwise general idea.