For the first part:
Assume $z=e^{it}$ as a substitution for the integral on the right side. Then
$\int_C \dfrac{i dz}{(z-1)(az-1)} = - \int_0^{2\pi} \dfrac{e^{it}dt}{a(1+e^{2it})-(1+a^2)e^{it}} = - \int_0^{2\pi} \dfrac{e^{it}dt}{a(1+\cos{2t}+i\sin{2t})-(1+a^2)(\cos{t}+i\sin{t})} = - \int_0^{2\pi} \dfrac{e^{it}dt}{(\cos{t}+i\sin{t})(2a\cos{t} -(1+a^2))} = \int_0^{2\pi} \dfrac{dt}{1+a^2 -2a\cos{t}}$.
For the second part,when 0≤a<1, the only singularity lies on the real axis within the unit circle at $a$ and it is a simple pole. So the residue can be computed as $\lim_{z\rightarrow a}(z-a)\dfrac{i}{(z-a)(az-1)}=\dfrac{i}{a^2−1}$.
Hence the integral is $2πi× Residue= \dfrac{2\pi}{1-a^2}$
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 $z$.
Best Answer
Define $g : C \times C^c \to \mathbb{C}$ by $$g(\zeta, z) = \frac{f(\zeta)}{(\zeta - z)^{n}}.$$ Let $g_z : C \times C^c \to \mathbb{C}$ denote the derivative of $g$ wrt $z$. Fix $z \in C^c$. You want to prove that $$\sup_{\zeta \in C}|\frac{g(\zeta, z + h) - g(\zeta, z)}{h} - g_z(\zeta, z)| \to 0 \text{ as } h \to 0.$$ By the FTC, for $\zeta \in C$, and $|h| < r := \text{dist}(z, C)$, $$\frac{g(\zeta, z + h) - g(\zeta, z)}{h} = \int_{0}^{1}g_z(\zeta, z + th)\,dt.$$ Hence for $\zeta \in C$ and $|h| < r$, \begin{align} |\frac{g(\zeta, z + h) - g(\zeta, z)}{h} - g_z(\zeta, z)| &= |\int_{0}^{1}(g_z(\zeta, z + th) - g_z(\zeta, z))\,dt| \\ &\leq \int_{0}^{1}|g_z(\zeta, z + th) - g_z(\zeta, z)|\,dt \\ &\leq \sup_{z_1 \in D_{|h|}(z)}|g_z(\zeta, z_1) - g_z(\zeta, z)| \\ &\leq \omega_{g_z, C \times \overline{D_{|h|}(z)}}(|h|), \end{align} where $\omega_{g_z, C \times \overline{D_{|h|}(z)}}$ is a modulus of continuity for the uniformly continuous function $g_z$ on $C \times \overline{D_{|h|}(z)}$. $g_z$ is indeed uniformly continuous on $C \times \overline{D_{|h|}(z)}$ because $g_z$ is continuous on $C \times C^c$, and $C \times \overline{D_{|h|}(z)}$ is compact. For $|h| < r/2$, $$\omega_{g_z, C \times \overline{D_{|h|}(z)}}(|h|) \leq \omega_{g_z, C \times \overline{D_{r/2}(z)}}(|h|) \to 0 \text{ as } h \to 0.$$ This completes the proof.
Edit: Here are some details on "modulus of continuity". I like to think of the "canonical" modulus of continuity of a function $f : X \to Y$ between two metric spaces as the map $\omega : [0, \infty) \to [0, \infty]$ defined by $$\omega(\delta) = \sup_{d(x_1, x_2) \leq \delta}d(f(x_1), f(x_2)).$$ Note that $d(x_1, x_2) \leq \delta \implies d(f(x_1), f(x_2)) \leq \omega(\delta)$. It is easy to show that $f$ is uniformly continuous if and only if $\omega(\delta) \searrow 0$ as $\delta \searrow 0$.
For this problem, we don't need to compute the modulus of continuity explicitly, i.e. we don't need hard quantitative estimates on it. For this problem, all was needed was that $g_z$ was continuous so that we could conclude that it's modulus of continuity has the limit $0$ at $0$.