Two possible definition of differentiability between two surfaces

The definition of a differentiable map between two surfaces given in the course is extrinsic. I explain it. Consider  $f:S_1\rightarrow S_2$ a map between two surfaces and $p\in S_1$. Then $f$ is differentiable at $p$ if $i\circ f\circ X:U\rightarrow{\mathbb R}^3$ is differentiable at $q=X^{-1}(q)$, where $i:S_2\rightarrow{\mathbb R}^3$ is the inclusion map (definition I). Here we use strongly that $S_2$ is included in Euclidean space ${\mathbb R}^3$. If one changes the viewpoint, one would request that the definition does not depend if $S_2$ is or is not included in ${\mathbb R}^3$, but only on $S_2$, that is, an intrinsic definition. Then the natural way to do it is by means of parametrizations and the definition would be: $f$ is differentiable at $p$ if $Y^{-1}\circ f\circ X:U\rightarrow W$ is smooth at $q=X^{-1}(p)$, where $X:U\rightarrow S_1$ and $Y_W\rightarrow S_2$ are parametrizations around $p$ and $f(p)$ respectively (definition II). Now it is not important if the surface is included in Euclidean space. 

This allows to extend the above definition to object with similar properties than surfaces, that is, objects with a set of parametrizations between open sets of ${\mathbb R}^n$ and open sets of the object. Then it will appear the concept of manifold of dimension $n$.

Returning, we prove that both definition are equivalents. 

1. (II) $\Rightarrow$ (I). Suppose a such $f$ which is differentiable at $p$ with definition II. When we consider a parametrization $X$ around $p$, then $i\circ f\circ X=(i \circ Y)\circ (Y^{-1} \circ f\circ X)$ and thus, it is the composition of two smooth maps between open sets of Euclidean spaces.
2. (I) $\Rightarrow$ (II). Suppose $f$ which is differentiable at $p$ with definition I, that is, we know $i\circ f\circ X$ is smooth at $q$ for any $X$. Without loss of generality, and fi $Y=(Y_1,Y_2,Y_3)$, we suppose that $$\left|\begin{array}{cc}\frac{\partial Y_1}{\partial u}&\frac{\partial Y_2}{\partial u}\\  \frac{\partial Y_1}{\partial v}& \frac{\partial Y_2}{\partial v}\end{array}\right|\not=0.$$ The Inverse function theorem asserts that the function $$(Y_1,Y_2):W'\rightarrow O', (u,v)\mapsto (Y_1(u,v),Y_2(u,v))$$ is a diffeomorphism between suitable open sets of ${\mathbb R}^2$. Let $\phi=(Y_1,Y_2)^{-1}$. If $(i\circ f\circ X)=(f_1,f_2,f_3)$, then $$Y^{-1} \circ f\circ X (u',v')=\phi^{-1} (f_1(u',v'),f_2(u',v'')),$$ which is differentiable because it is the composition of two differentiable maps.

Finally, we will adopt the definition I because it is more intuitive, although we are loosing `generality'.