1) First a geometric description: For $\lvert \tau\rvert = 1$ we have $1/\tau = \overline{\tau}$ (off the unit circle there's also some scaling involved). Thus when $\tau$ travels along the unit circle from $\rho$ to $i$, then $1/\tau$ travels along the unit circle from $\overline{\rho} = \rho^2$ to $\overline{i} = -i$, and therefore $-1/\tau$ travels along the unit circle from $-\overline{\rho} = -\rho^2 = \rho + 1$ to $-\overline{i} = -(-i) = i$.
Let's do it with a parametrisation: Since $\rho = \exp \bigl(i\frac{2\pi}{3}\bigr)$ and $i = \exp \bigl(i\frac{\pi}{2}\bigr)$ we can parametrise the arc from $\rho$ to $i$ by $\tau(t) = \exp\bigl(i\bigl(\frac{2\pi}{3} - t\bigr)\bigr)$, $0 \leqslant t \leqslant \frac{\pi}{6}$. Then
$$-\frac{1}{\tau(t)} = -\frac{1}{\exp\bigl(i\bigl(\frac{2\pi}{3} - t\bigr)\bigr)} = -\exp\bigl(i\bigl(t - \tfrac{2\pi}{3}\bigr)\bigr) = \exp\bigl(i\bigl(t - \tfrac{2\pi}{3}\bigr) + i\pi\bigr) = \exp\bigl(i\bigl(\frac{\pi}{3} + t\bigr)\bigr)\,.$$
We see that this is an arc on the unit circle too, it starts at $-1/\tau(0) = \exp\bigl(i\frac{\pi}{3}\bigr) = \rho + 1$ and it ends at $-1/\tau(\pi/6) = \exp\bigl(i\bigl(\frac{\pi}{3} + \frac{\pi}{6}\bigr)\bigr) = \exp\bigl(i\frac{\pi}{2}\bigr) = i$. The original arc was traversed in direction of decreasing argument (i.e. clockwise), while the transformed arc is traversed in direction of increasing argument (counterclockwise), that is, the orientation is reversed.
2) Let us split the fundamental region $R_{\Gamma}$ in two parts, $A_M$ shall be the part where $\operatorname{Im} \tau < M$, and $B_M$ the part where $\operatorname{Im} \tau > M$, where $M$ is chosen large enough that $f$ has neither zeros nor poles in $B_M$. ($B_M$ shall not contain $i\infty$.) Let $N_M$ be the number of zeros of $f$ in $A_M$, and $P_M$ the number of poles of $f$ in $A_M$. Then by the argument principle
$$N_M - P_M = \frac{1}{2\pi i} \int_{\partial A_M} \frac{f'(\tau)}{f(\tau)}\,d\tau\,.$$
By the preceding discussion the integrals over the two vertical segments of the boundary cancel, and the integrals over the two arcs on the unit circle cancel too, thus only the integral over the horizontal line remains,
$$N_M - P_M = \frac{1}{2\pi i} \int_{\frac{1}{2} + iM}^{-\frac{1}{2} + iM} \frac{f'(\tau)}{f(\tau)}\,d\tau\,.$$
Now we express this integral in terms of $x = e^{2\pi i\tau}$. Since the real part $u$ of $\tau = u + iM$ decreases in that integral, the circle $\lvert x\rvert = e^{-2\pi M}$ is traversed clockwise, i.e. in the negative direction (hence the minus sign in the next formula). Thus
$$N_M - P_M = -\frac{1}{2\pi i} \int_{\lvert x\rvert = e^{-2\pi M}} \frac{F'(x)}{F(x)}\,dx\,.$$
By the argument principle, and taking the sign into account, this is $P_F - N_F$ and we obtain
$$N_M - P_M = P_F - N_F$$
or after rearranging
$$N_M + N_F = P_M + P_F\,.$$
But $N_M + N_F$ is the total number of zeros of $f$ in $R_{\Gamma}$ (including a possible zero at $i\infty$) and $P_M + P_F$ is the total number of poles of $f$ in $R_{\Gamma}$ (including a possible pole at $i\infty$). So overall $f$ has the same number of zeros and poles in $R_{\Gamma}$, when we include $i\infty$.
Apostol's $N$ and $P$ — my $N_M$ and $P_M$ — are the number of zeros and poles in $R_{\Gamma}$ excepting $i\infty$, whereas the total number of zeros and poles must include $i\infty$. I think that's what threw you off.
3) I hope that's clarified under 2), otherwise I don't understand the problem here. (If not, please explain the problem further, I'll be happy to elaborate.)
4) The circular arc $C_1$ goes from the vertical line $\operatorname{Re} \tau =-\frac{1}{2}$ to the unit circle. Its centre is $\rho$. When the radius $r$ is small, the endpoint of $C_1$ on the unit circle is very close to the intersection of $C_1$ with the tangent to the unit circle at $\rho$. The tangent has the parametrisation $\rho - t\cdot i\rho$, $t \in \mathbb{R}$. The angle it forms with the real axis is the argument of its direction $-i\rho = \exp(-i\pi/2)\exp(2\pi i/3) = \exp(\pi i/6)$. Hence the angle between the vertical line and the tangent is $\frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$. Since the unit circle lies completely below the tangent (excepting the point of contact $\rho$), the angle subtended by $C_1$ is always larger than $\pi/3$. But rotating the tangent clockwise by an angle $\varepsilon > 0$ produces a secant of the circle, and for all sufficiently small $r$, such that the endpoint of $C_1$ lies between $\rho$ and the other intersction of the secant and the unit circle, the angle subtended by $C_1$ lies betwen $\pi/3$ and $\pi/3 + \varepsilon$. Since $\varepsilon$ can be arbitrarily small it follows that the limit as $r \to 0$ of the angle subtended by $C_1$ is $\pi/3$.
I believe your confusion in 1 is you are switching variables in your head. We want to show for fixed $a$, $I(s,a)$ is entire in $s$. However, the curves $C_1,C_2$ and $C_3$ are defined in the $z$ variable. So you don't need to show uniform convergence for small disks as a separate case. There is no direct interplay between the disk in $s$ and the curves defined in $z$. So, if you show uniform convergence for arbitrarily large disks in the $s$ variable, you are finished.
Implicitly, the author thinks of $I(s,a)=I_1(s,a)+I_2(s,a)+I_3(s,a)$ where $I_j$ is integrating over $C_j$. To the author , it is clear $I_2$ is entire as the integrand is holomorphic in $z$ in some domain containing $C_2$.
Proving 2 for finite length curves is tantamount to proving you can differentiate under an integral in appropriate circumstances. For curves of infinite length, you need to do some work involving bounds and uniform convergence. (as the author does).
Suppose $f(s,z)$ is defined on $U\times V$ and is holomorphic in either variable while the other is fixed. Let $C$ be a finite length curve in $V$. Then $I(s)=\int_C f(s,z) dz$ is holomorphic and $I'(s)=\int_C f_s(s,z) dz$. Proof: differentiate under the integral sign.
Best Answer
$T$ isn't being removed. Nor is it tending to anything. Only three facts are being used in going from the line above "Let $b\to\infty$": firstly, that $\log(a) = -\log(1/a)$ (to get rid of the minus sign on the top of the first fraction), secondly, that $\frac{2Ta^b}{b}\to 0$ as $b\to \infty$ (to get rid of the second fraction), and thirdly the fact that we can pass from $A \leq B(b)$ for all $b$ to $A \leq \lim_{b\to\infty}B(b)$.
To derive the initial inequality, we have
\begin{align*}\left|\int_{c-iT}^{c+iT}a^z\frac{dz}{z}\right| &= \left|\int_{b+iT}^{c+iT}a^z\frac{dz}{z} + \int_{b-iT}^{b+iT}a^z\frac{dz}{z} + \int_{c-iT}^{b-iT}a^z\frac{dz}{z}\right|\\ &\leq \left|\int_{b+iT}^{c+iT}a^z\frac{dz}{z}\right|+\left|\int_{b-iT}^{b+iT}a^z\frac{dz}{z}\right|+\left|\int_{c-iT}^{b-iT}a^z\frac{dz}{z}\right|\\ \end{align*}
So now we just need to show that the middle term is bounded above by $\displaystyle\frac{2Ta^b}{b}$, while the other two are each bounded above by $$\int_{b}^c\frac{a^x}{T}dx.$$ We start with the former: \begin{align*}\left|\int_{b-iT}^{b+iT}a^z\frac{dz}{z}\right| &= \left|\int_{-T}^T\frac{a^{b+ix}}{b+ix}dx\right|\\&\leq \int_{-T}^T\left|\frac{a^{b+ix}}{b+ix}\right|dx\\&= \int_{-T}^T\frac{a^b}{|b+ix|}dx\\&\leq \int_{-T}^T\frac{a^b}{b}dx\\&=\frac{2Ta^b}{b}.\end{align*} Now, the first and last summand: \begin{align*} \left|\int_{b\pm iT}^{c\pm iT}a^z\frac{dz}{z}\right| &= \left|\int_{b}^c\frac{a^{x\pm iT}}{x\pm iT}dx\right|\\&\leq \int_b^c\left|\frac{a^{x\pm iT}}{x\pm iT}\right|dx\\ &=\int_b^c\frac{a^x}{|x\pm iT|}dx\\ &= \int_{b}^c\frac{a^x}{T}dx, \end{align*} as required.