What does $f$ is holomorphic on the neighborhood of the unit disc mean? Does it mean it's holomorphic on the disc or both on&inside the disc?
Also, if $f$ is holomorphic on the neighborhood of the unit disc, why is the Cauchy Theorem leads from the first line to the second line?
It would be great if someone could help me out :) Thanks!
Generally when people say that a function $f$ is holomorphic on a neighborhood of the unit disk, they mean that there is an open set $\Omega$ containing the disk $\{z:|z|\leq 1\}$ such that $f$ is holomorphic on $\Omega$. Often $\Omega$ is assumed to be connected as well.
To pass from the first line to the second in the equations you included, we must show that $$ \int_{|z|=1}\frac{f(z)}{z-\overline{z}_0^{-1}}\;dz=0 $$
Although you didn't specifically say this, my guess is that $0<|z_0|<1$. Then $|\bar{z}_0^{-1}|=\frac{1}{|z_0|}>1$, and it follows that the function $\frac{f(z)}{z-\overline{z}_0^{-1}}$ is holomorphic in a neighborhood of the closed unit disk. Therefore $$ \int_{|z|=1}\frac{f(z)}{z-\overline{z}_0^{-1}}\;dz=0 $$ by Cauchy's theorem.