While in the first part of the book (until chapter 4), one gets by without differential forms, in this part they become quite useful. In fact, Chapter 5 goes full on into differential form mayhem even on the first page :), but there it is totally unavoidable. I am currently writing up some appendixes for the book and one of them will be a very short (without proofs) review of differential forms and Stokes theorem. Though that's not done yet. Best is to look through the chapter on differential forms in a book like baby Rudin.
(1) Yes, it is the differential form version of the Stokes theorem. Really in this setting it is the Green's theorem (which is Stoke's theorem in two variables, or perhaps, classical Stokes in the $xy$-plane). As to why we take out a small disc, that is because the function that we would want to integrate over $U$ has a singularity in $U$, so Stokes does not apply. So we take out a small disc around the singularity. Then the theorem does apply since everything is nicely $C^1$ on the $U \setminus \Delta_r$.
It is not a bad exercise to use the classical Green's theorem formulation and see if you can make the proof work. That is, use Green's theorem and then see if you can end up with the second integral in (**) using the classical formulation. It is slightly more tedious to write down that way, but it is not that terrible.
(2) For differential one-forms, $dx \wedge dy = - dy \wedge dx$ (let's say $x$ and $y$ are two coordinates). This means that $dx \wedge dx = -dx \wedge dx = 0$. Same thing works for the complex forms $dz$ and $d\bar{z}$ instead of $x$ and $y$ (if $z=x+iy$, then $dz = dx + i \,dy$, and $d\bar{z} = dx - i\,dy$). Then the equality just follows from the definition of the $d$ operator on differential forms:
$d(a\, dz + b\, d\bar{z}) =
\frac{\partial a}{\partial z} dz \wedge dz +
\frac{\partial a}{\partial \bar{z}} d\bar{z} \wedge dz +
\frac{\partial b}{\partial z} dz \wedge d\bar{z} +
\frac{\partial b}{\partial \bar{z}} d\bar{z} \wedge d\bar{z}
=
\frac{\partial a}{\partial \bar{z}} d\bar{z} \wedge dz +
\frac{\partial b}{\partial z} dz \wedge d\bar{z}
=
\left(\frac{\partial b}{\partial z}-\frac{\partial a}{\partial \bar{z}} \right) d\bar{z} \wedge dz$
(3) In the first guy on the left in (**) that is just $d\zeta$ (no wedge there) that's evaluated over the boundary of the disc. This is just the normal path integral from one variable complex analysis. In fact, this is precisely the same argument from the standard proof of Cauchy formula that is usually given (you reduce to a disc centered at a point). The idea is to now actually compute the integral, so parametrize by saying that $\zeta = z+re^{it}$, then $\zeta-z = re^{it}$ and $d\zeta = rie^{it}dt$. The denominator gets canceled, the $i$ also gets canceled, and you get the integral in the middle. Now this is where you use continuity, the integral is just the average of the value of $f$ over a tiny circle around $z$, so as $r \to 0$, the limit must be the value of $f$ at $z$.
This is a shameless adaptation from my own set of notes, but it does help (me) tie a few things together mentioned in the relevant PDF linked below the question.
Consider the principal value integral \begin{equation}\label{principalval}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} \:,\end{equation} where by shifting $x \rightarrow z$, we assume that $f\left(z\right)$ is analytic except for a finite number of poles, and that $\left|f\right| \rightarrow 0$ on the upper (or lower) infinite semicircle in the complex plane.
Since the pole $x_0$ lies on the real axis, the integration contour cuts directly through $x_0$. This is handled by regularization of the denominator, which entails introducing a small factor $\delta>0$ as $$I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \lim_{\delta\rightarrow 0} \int_{-\infty}^\infty \frac{\left(x-x_0\right)f\left(x\right) \: dx}{\left(x-x_0\right)^2 + \delta^2} = \lim_{\delta\rightarrow 0} \oint_C \frac{\left(z-x_0\right)f\left(z\right) \: dz}{\left(z-x_0\right)^2 + \delta^2} \:.$$ After a little complex algebra, find $$I = \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} + \lim_{\delta\rightarrow 0} \oint_C i\delta \frac{f\left(z\right) \: dz}{\left(z-x_0-i\delta\right)\left(z-x_0+i\delta\right)} \:,$$ which indicates one simple pole $z_0 = x_0 + i\delta$ inside the upper-half plane. The first integral in fact excludes the pole, so $x_0$ is skipped in subsequent residue calculations. (Use the $\delta$-term as a reminder to skip $x_0$.) The second integral is solved by standard residue calculus, i.e., let $g\left(z\right)=f\left(z\right)/\left(z-x_0+i\delta\right)$, resulting in $\pi i f\left(x_0\right)$.
Pulling the results together, we write \begin{equation}\label{principalvaltwoterms}I^+ = \pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0+i\delta} \:,\end{equation} where if we started with $\delta<0$ instead, the integration contour would flip to the lower-half plane, resulting in $$I^- = -\pi i f\left(x_0\right) + \lim_{\delta\rightarrow 0} \oint_C \frac{f\left(z\right) \: dz}{z-x_0-i\delta} \:.$$ In tighter notation (regardless of path or the sign of $\delta$), one may write \begin{equation}\label{principalvalcomplex}I = \text{P} \int_{-\infty}^\infty \frac{f\left(x\right) \: dx}{x-x_0} = \text{P} \oint_C \frac{f\left(z\right) \: dz}{z-x_0} \:,\end{equation} reminding us to include $x_0$ inside integration contour, but take the residue with a factor of $1/2$.
So for instance, to evaluate $$I = \int_{-\infty}^\infty \frac{\sin x}{x} \: dx \:,$$ the answer is almost too trivial to show off the method, but here it is:
$$I = \text{Im} \left( \pi i \: e^{i \cdot 0} + \lim_{\delta\rightarrow 0} \oint_C \frac{e^{iz} \: dz}{z+i\delta} \right) = \text{Im} \left( \pi i \: e^{i \cdot 0} + 0 \right) = \pi$$
By the same apparatus, we can show the cosine-version to be zero.
Best Answer
Yes, it's due to dominated convergence. For simplicity in notation, suppose $w=0$. The function $f$ is assumed $C^1$ on $K$, so $\frac{\partial f}{\partial \overline{z}}$ is bounded on $K$, so doesn't affect convergence. You are right to be worried because $x\mapsto \frac{1}{x}$ is not integrable near the origin in 1-dimension. The nice thing is that you're now in 2-dimensions, so although you still have the reciprocal denominator $\frac{1}{z}$, the difference is that you're integrating the form $\frac{1}{z}\,dz\wedge d\overline{z}$. In other words ignoring constant factors, the question is whether or not the integral \begin{align} \int_D\frac{1}{|(x,y)|}dx\,dy \end{align} is finite (here $D$ is the unit disc centered at the origin). By changing variables to polar coordinates, we see that \begin{align} \int_D\frac{1}{|(x,y)|}dx\,dy&=\int_0^1\int_0^{2\pi}\frac{1}{r}r\,d\theta\,dr=2\pi<\infty. \end{align} So, although $\frac{1}{z}$ has a singularity at the origin in $\Bbb{C}$, it is an integrable singularity with respect to the 2-dimensional Lebesgue measure (because $dz\wedge d\overline{z}\sim dx\wedge dy=r\,dr\wedge d\theta$). Just FYI: in $n$-dimensions, $\int_{B_{1}(0)}\frac{1}{\|x\|^s}\,d^nx$ is finite if and only if $(n-1)-s>-1$, i.e if and only if $s<n$ (in the case above, $s=1$ and $n=2$).
Putting this all together, $\frac{1}{(z-w)}\frac{\partial f}{\partial \overline{z}}\frac{i}{2}\,dz\wedge d\overline{z}$ is a $L^1_{\text{loc}}(\Bbb{C})=L^1_{\text{loc}}(\Bbb{R}^2)$ (locally integrable) differential form in the sense that we can write it as $\phi(x,y)\,dx\wedge dy$ for some $\phi$ whose integral over compact sets is finite. So, it is valid by Dominated convergence ($|\phi|$ serves as the dominating function) to first remove an $\epsilon$ ball around the point $w$, calculate integrals, and finally take limits.