What is the name of this multivariable derivative? This is (hopefully) the definition of a tangent vector based on curves (M is a manifold):
$$
\lim_{t\to 0}\frac{[\gamma;\phi;\phi^{-1};f](0 + t\cdot1) - [\gamma;\phi;\phi^{-1};f](0)}{t}
$$
where $\gamma\colon \mathbb{R}\to M$, $\phi\colon M\to\mathbb{R}^{k}$, $\phi^{-1}\colon \mathbb{R}^{k}\to M$, $f\colon M\to\mathbb{R}$ (used ; for diagrammatic composition).
This made me wonder: why only one parameter? Is there any derivative that is like this:
$$
\lim_{t\to 0}\frac{[\Gamma;\phi;\phi^{-1};F](0 + t\cdot X) - [\gamma;\phi;\phi^{-1};F](0)}{t}
$$
with $\Gamma\colon \mathbb{R}^m\to M$ and $F\colon M\to \mathbb{R}^n$, and what is it called in that case? 
(Guess: maybe this variant is used for maps between manifolds? I just remember a lot of talk about "implicit function theorem" when reading about that...)
 A: I hope this will clarify things a little.
In more typical notation $[\Gamma;\phi;\phi^{-1};F]=F\circ\phi^{-1}\circ\phi\circ\Gamma$.
Since $\phi^{-1}\circ\phi\colon M\to M$ is the identity, we have in fact $F\circ\phi^{-1}\circ\phi\circ\Gamma=F\circ\Gamma=:G$.
This $G$ is a map $\mathbb R^n\to\mathbb R^n$, so you can define its derivative in the usual Euclidean way.
It should be a familiar result from multivariable calculus that
$$
\lim_{t\to0}\frac{G(x+tX)-G(x)}{t}
=
DG(x) X
$$
for any $x\in \mathbb R^n$ and $X\in\mathbb R^n$.
Here $DG$ is the derivative matrix of $G$.
This is the directional derivative in the direction of $X$ at the point $x$.
Notice that both $F$ and $\Gamma$ require coordinates on the manifold to describe and differentiate, but the composition does not1.
The composed function is an honest function from a Euclidean space to another one.
If you differentiate using the chain rule, then coordinates come into play, although it can be more convenient to write things in an invariant form.
Your first derivative can be written as $df(\dot\gamma)$.
Here the differential $df$ is in the cotangent space $T_{\gamma(0)}^*M$ and $\dot\gamma(0)\in T_{\gamma(0)}M$.
This makes $df(\dot\gamma)$ naturally a scalar.
In the second case $dF_{\Gamma(0)}$ is a linear map $T_{\Gamma(0)}M\to T_{F(\Gamma(0))}\mathbb R^n=\mathbb R^n$.
The chain rule says $DG(x)=dF_{\Gamma(x)}\circ d\Gamma_x$.
Now $x=0$, so the derivative you want to calculate is
$$
DG(0)X
=
dF_{\Gamma(0)}\circ d\Gamma_0 X.
$$
Now $d\Gamma_0 X$ is a tangent vector at $\Gamma(0)$.
You can often study a function $G$ from anywhere to $\mathbb R^n$ component by component.
It makes no real difference that the target is multidimensional.
Also, assuming everything is nice and smooth, the derivative only depends on $\Gamma$ restricted to the one-dimensional subspace $X\mathbb R\subset\mathbb R^n$.
So from a certain point of view having $n>1$ has little effect on the differential geometry here.

1
If a function is defined to or from a manifold, you can always use the invariant objects to differentiate them. But often it is best to use some local coordinates on the manifold. When you do so, you get a coordinate representation of the derivative. The differentials $dF$ and $dΓ$ will depend on the choice of coordinates, but the differential of $G$ will not.
