My question is how to get from $$c_n=\frac{1}{2\pi i} \int_{|z-z_0|=\rho}\frac{f(z)}{(z-z_0)^{n+1}}dz$$ so the usual formula for Coeffizients for Laurentseries to the integral representation of the fourier coefficients of Modular forms, namely $$c_n=\int_0^1 f(\tau)\cdot e^{-2\pi n \tau i}d\tau.$$
I have not found a rigorous solution. I thought about taking somehow $\frac{1}{z}$ and compute the integral of this over the unit circle which gives $2\pi i$. So we could cancel $\frac{1}{2\pi i}$. Now one could somehow use the fact that for a modular form we have $f(z)=\hat{f}(q)$ where $q=e^{2\pi i z}$. Thats would explain the $e^{-2\pi n \tau i}$. However, I am not sure either how to get the wrong bounds of the integral.
If we start with $$\frac{1}{2\pi i} \int_{|z-z_0|=\rho}\frac{f(z)}{(z-z_0)^{n+1}}dz$$ with $z_0=0$ and $0<\rho<1$, and set $z=\rho e^{2\pi ix}$, then the result is $$ \int_0^1f(\rho e^{2\pi ix})\rho^{-n}e^{-2\pi inx}\;dx $$ Since $0<\rho<1$, we can write $\rho=e^{-2\pi y}$ for some $y>0$, so setting $\tau=x+iy$ the above becomes $$ \int_0^1f(e^{2\pi i\tau})e^{-2\pi in\tau}\;d\tau$$
In particular, if $f$ is defined by $f(e^{2\pi i\tau})=g(\tau)$ for a modular form $g$, then this becomes $$ \int_0^1g(\tau)e^{-2\pi in\tau}\;d\tau$$
Essentially what's going on is that the map $\tau\mapsto e^{2\pi i\tau}$ takes the upper half plane to the punctured unit disk, with lines of constant imaginary part mapping to circles centered at the origin.