I don't quite understand the proof of Stokes' theorem. So the Stokes' theorem says $$\oint_C \mathbf F\cdot d\mathbf r = \iint_S (\nabla\times\mathbf F) \cdot d\mathbf s$$
In the proof it says we can divide $S$ into smaller pieces $S_1, S_2, ... $ and divide $C$ into $C_1, C_2, ...$ . I can understand until here. But when representing surface $S$, I don't understand why $S$ can be represented by $$z = h(x,y)$$ What if $z$ is multi-valued for each $x$ and $y$ ? Then how do we prove the theorem?
There is no simple proof of Stokes' theorem $$\int_{\partial S}{\bf F}\cdot d{\bf r}=\int_S(\nabla\times{\bf F})\cdot d{\bf s}\ .\tag{1}$$ The proof in your book can be salvaged as follows: The given "large" surface $S$ can be partitioned into finitely many small surfaces $$S_i:\quad z=h_i(x,y)\qquad\bigl((x,y)\in D_i\bigr)\qquad(1\leq i\leq N)\ ,\tag{2}$$ whereby some of the $S_i$ may also be in the form $x=h_i(y,z)$ or $y=h_i(x,z)$. You then prove Stokes' theorem for an $S_i$ of the fom $(2)$, whence it is also proven if $S_i$ is of one of the other two forms. Putting it all together we obtain $(1)$ since the integrals along inner dividing curves separating the $D_i$ cancel.