In order to better understand Green's Theorem, I developed this informal proof, which I request verification and critique of (both the proof and its writing). Of course, any textbook has a proof: my goal here is to improve my intuition by developing a proof from a simple argument I could visualize, and then passing to the limit.
Consider a linear function $\mathbf F: \mathbb R^2 \to \mathbb R^2$. Write $\mathbf{F}(x,y) = P(x,y) \mathbf i + Q(x,y) \mathbf j$, and let $p = P(0,0)$ and $q = Q(0,0)$. Then, for any rectangle $C$ parallel to the $x$ and $y$ axes with width $w$ and height $h$, $$ \frac 1 {hw} \oint_C \mathbf F \cdot d\mathbf l$$ is the circulation of $\mathbf F$ around $C$ divided by the area of $C$.
Define the 2D curl of $\bf F$ at $(x,y)$ as $$\lim_{h,w \to 0} \frac 1 {hw} \oint_C \mathbf F \cdot d\mathbf l$$ for any rectangle $C$ parallel to the $x$ and $y$ axes with width $w$ and height $h$, provided that this quantity exists and is identical for any sequence of such rectangles each containing $(x,y)$.
We claim that the 2D curl of $\mathbf F$ is $\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y}$. For consider any rectangle $C$ with corners $(-u, -v), (u, -v), (u,v), (-u,v)$. Then, $hw = 4uv$ and, since $\mathbf F$ is linear, \begin{align} \oint_C \mathbf F \cdot d \mathbf l &= 2u\left(p - \frac {\partial P}{\partial y}v\right) + 2v\left(q + \frac {\partial Q}{\partial x}u\right) - 2u\left(p + \frac {\partial P}{\partial y}v\right) - 2v\left(q - \frac {\partial Q}{\partial u}v\right) \\ &= 4uv\left(\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y}\right). \end{align} (This is the crux of the argument. The rest is simply set up or justification of using the linear approximation to find the limit.)
If we relax the assumption that $\mathbf F$ is linear, but still assume it to be smooth then over a sufficiently small rectangle, the linear approximation holds to arbitrary accuracy, and the limit still holds. So the 2D curl of $\mathbf F$ at $(x,y)$ is $\frac {\partial Q}{\partial x}(x,y) - \frac {\partial P}{\partial y}(x,y)$.
Now consider an arbitrary smooth region $R$. It can be shown via a simple geometric argument that the circulation around $R$ is equal to the sum of the circulations around any decomposition of $R$. Then for any grain size $G > 0$, we can decompose $R$ into rectangles parallel to $x$ and $y$ axes of height and width $G$, plus a peripheral region. As $G \to 0$, the area of this periphery shrinks faster than $G$, and since $F$ is assumed smooth, the circulation of the periphery goes to zero as well. Thus, as $G \to 0$, the sum of the circulation around all of the rectangles goes to the circulation around the entire region. But the sum of the circulation around all of the rectangles is simply $\iint_R \frac {\partial Q}{\partial x}(x,y) - \frac {\partial P}{\partial y}(x,y) \, dx \, dy$, giving as desired $$\oint_{\partial R} \mathbf F \cdot d\mathbf l = \iint_R \frac {\partial Q}{\partial x}(x,y) - \frac {\partial P}{\partial y}(x,y) \, dx \, dy.$$