let $\mathcal{S},\mathcal{\tilde{S}}$ be two surfaces, and let $f$ be a smooth map: $$ f:\mathcal{S}\rightarrow\mathcal{\tilde{S}}. $$ If $p\in\mathcal{S}$ and $\tilde{p}\in\mathcal{\tilde{S}}$, then the derivative $D_p$ of $f$ is defined as a map:
$$ D_p f : T_p \mathcal{S} \to T_{f(p)}\mathcal{\tilde{S}}. $$
Specifically the Gauss map is a map that goes from $\mathcal{S}$ to $S^2$ (the unitary sphere). And the Weingarten map is defined as
$$ \mathcal{W}_{p,\mathcal{S}} = -D_p\mathcal{G} $$
My question is... since there's a definition of derivative, what's the definition of differential? And how can we express the differential of $\mathcal{G}$?
The OP has asked for a more in-depth explanation of tangent vectors, although it is a slight digression from the original question. However, I cannot fit an explanation in comments. Hence the answer.
Let $\phi: I \rightarrow M$ be a curve parameterized by some interval such that $x = \phi(0)$. $x \in M$ is the point we are interested in. There is some chart $(U, \psi)$ such that $x \in U$ and $U \subset M$. There is no notion of derivative on a bare manifold, and thus we need to make use of $\psi$. Specifically, we are interested in the map $\psi \circ \phi: I \rightarrow \Bbb{R}^n$. Now we do have a well-defined notion of tangent vector with the map $\psi \circ \phi$ which corresponds to the notion of $\mathbf{T}$ that you learned in multivariable calculus.
The tangent vector at $x$ is then $\frac{d}{dt} \psi \circ \phi$ evaluated at $\psi(\phi(0))$. The tangent space $T_xM$ arises when you consider all possible curves through $x$, at all possible parameterizations of those curves.