I am looking at the proof of the Green theorem.
To show that $$\oint _S (Mdx+Ndy)= \iint_R \left( \frac{\partial{N}}{\partial{x}}-\frac{\partial{M}}{\partial{y}} \right) dxdy$$we do the following:
To show that $\oint _S Mdx=-\iint_R \frac{\partial{M}}{\partial{y}}dxdy$,we consider this graph:
And to show that $\oint _S Ndy=\iint_R \frac{\partial{N}}{\partial{x}}dxdy$,we consider this graph:
Why do we choose these graphs at each case?
(1) Line integral is independent of parametrization :
If $f = (P,Q)$ is a vector field, then assume that a curve $C$, a boundary of $R$, has a parametrization $r(t)=(x,y)(t),\ 0\leq t\leq l$. Then for a line integral $ \int_C f\cdot dr $ we have to consider $\ast=\int_0^l P(x(t),y(t)) x'(t) dt$
Here $s(u)=r(t(u)),\ t(a)=0,\ t(b)=l$ is a re-parametrization so that $$ \ast =\int_a^b P(x(t(u)),y(t(u)) ) x'(t(u)) t'(u) du = \int_a^b P(s(u)) \frac{d}{dt} x( t(u)) du $$
Hence $\ast$ is independent of parametrization
(2) Proof of Green theorem :
For $\ast$, we use the following specific parametrization $c_1(t)=(t,f_1(t)),\ a\leq x\leq b$ and $c_2(t) =(b-t,f_2(b-t)),\ 0\leq t\leq b-a$ for $S$ (cf. first figure in OP)
\begin{align*}&\int_{c_1\cup c_2} P(x,y)dx \\&= \int_a^b P(t,f_1(t)) dt + \int_0^{b-a} P(b-t,f_2(b-t)) \frac{d}{dt} (b-t) dt \\&= \int_a^b P(t,f_1(t)) dt + \int_b^{a} P(u,f_2(u)) du \\&=\int_a^b \bigg\{P(t,f_1(t)) - P(t,f_2(t)) \bigg\} dt \\&=\int_a^b \int_{f_1(t)}^{f_2(t)} -P_y(t,y) dydt \\&= \int\int_R -P_y dA \end{align*}
And for $\int_C Q(x,y)dy$, if we use $c_1,\ c_2$, we can not calculate easily as far as I know. So we use another parametrization $c_1=(g_2(y),y),\ d\leq y\leq c$ and $c_2(t)=(g_1(c-t),c-t),\ 0\leq t\leq c-d$.