I can understand your confusion, I think beginners in the subject are often confused by the amount of identifications and subtleties which come up in explicit calculations. Your professor probably just doesn't see these anymore. I hope I can clear some of those for you.
Normally, derivatives of maps between manifolds are computed explicitly using their representation in some charts, e.g. if $f: M^m\rightarrow N^n$ and $\varphi_1: U_1\rightarrow V_1\subset \mathbb{R}^m$ and $\varphi_2: U_2\rightarrow V_2\subset \mathbb{R}^n$ are charts of $M$ and $N$, respectively, we compute the usual vector space derivative (the Jacobian) of $\varphi_2 \circ f \circ \varphi_1^{-1}: V_1\rightarrow V_2$.
I suppose in your post $S$ is the 2-sphere and your parametrization is given as a map $\mathbb{R}^2\rightarrow\mathbb{R}^3$. So implicitely, for your first manifold $M=\mathbb{R}^2$ there is already a chart chosen, namely the identity chart $\varphi_1=\operatorname{id}_{\mathbb{R}^2}$.
For the second manifold $N=S$ there is no chart chosen. Instead the 2-sphere is seen as a subset\submanifold of $\mathbb{R}^3$ by the use of the embedding $\iota: S\rightarrow \mathbb{R}^3$. Using the said above, the actual manifold map $f:\mathbb{R}^2\rightarrow S$ is identified with $\psi=\iota\circ f \circ\operatorname{id}_{\mathbb{R}^2}^{-1}$, which is exactly the given parametrization.
So you have a map $\mathbb{R}^2\rightarrow\mathbb{R}^3$ for which you have a precise idea what a derivative is, compute it! The above identification then relates it to the derivative of $f$:
$$
d\psi = d\iota \circ df \circ d\operatorname{id}^{-1}.
$$
Since all of these steps are somehow canonical, in practice $d\psi$ and $df$ are considered to be "the same".
