How to show that $f(z)= 1/(z-a)$ is periodic with period $2\pi i$?

Ahlfors is talking about a different kind of period there, which are not the same as periods in the definition of a periodic functions. Indeed, on page 147 Ahlfors defines periods to be the values of the contour integrals $$\int_{\gamma_i}f\,dz,$$ where $\gamma_i$ are a homology basis of the domain. In the case of the domain $\Omega'=\{0<|z-a_j|<\delta_j\}$ discussed in the cited passage, homology basis has precisely one element $\gamma$, which is a counterclockwise loop $\gamma$ going once around $a_j$ contained in $\Omega'$ (we would also take a clockwise loop, which would change the sign of the result, but nothing else), and therefore we have exactly one period of $f(z)=\frac{1}{z-a_j}$, namely $$\int_\gamma\frac{dz}{z-a_j}=2\pi i.$$

I would also like to mention that the terminology, while perhaps confusing, is not unjustified. There is a general theory explaining it, but let me explain it in a special case of a function $\frac{1}{z}$.

As you probably know, if we fix a base point $1$, then for $z_0\neq 0$ the value of an integral $$\int_1^{z_0}\frac{dz}{z}$$ is not well-defined - indeed, it depends on the choice of the path from $1$ to $z_0$. However, we can consider a "multivalued function" which arises this way, by considering all possible values. This, as you can easily figure out yourself, is the logarithm function $\log z$. Moreover, while this function is not well-defined, its inverse function, $e^z$, is, and moreover it is, surprise, periodic with period $2\pi i$! You can actually figure that out directly from the fact that the integral over a circle is equal to $2\pi i$.

In general this is more complicated, because you have to consider integrals of multiple functions (or, more precisely, differential forms); essentially, this means that you have to choose a basis of not just homology, but also cohomology. Nevertheless, the idea is essentially the same.

It is not periodic in that sense. By period he means the integral over a small circle about $a$, and you know this is $2\pi i$.