There are many equivalent definitions of Teichmuller space for a surface of genus $g\ge 2$.
One of them concerns the complex structure: the Teichmuller space $\mathcal{T}(g)$ is the set of the couples $(X,f)$ where $X$ is a Riemann surface of genus $g$ and $f:S\rightarrow X$ is the marking, i.e. an orientation preserving diffeomorphism up to the equivalence $(X,f)\sim(X',f')$ iff there is a biholomorphism homotopic to $f'\circ f^{-1}$.
Another definition concerns the hyperbolic structure: $\mathcal{T}(g)$ is the set of the couples $(Y,h)$ there $Y$ is a hyperbolic surface of genus $g$ and $h:S\rightarrow Y$ is an orientation preserving diffeomorphism up to the equivalence $(Y,h)\sim(Y',h')$ iff there is an isometry homotopic to $h'\circ h^{-1}$.
My question is: how are these two definitions equivalent? How can I explicitly obtain the complex structure from the hyperbolic one and vice versa?
Obtaining a complex structure from a hyperbolic structure is quite easy. The hyperbolic plane (using either the Poincare disk or upper half plane model) can be viewed as a subset of the complex plane, and orientation-preserving hyperbolic isometries are biholomorphic. Thus any atlas of orientation-preserving hyperbolic charts on an oriented hyperbolic surface can also be viewed as an atlas of complex charts, so any oriented hyperbolic surface is automatically a complex surface.
The reverse direction is a little bit harder. If $X$ is a Riemann surface (i.e. a surface with a complex structure), then the universal cover $\widetilde{X}$ of $X$ must be a simply connected Riemann surface. According to the Uniformization Theorem, the only simply connected Riemann surfaces are the Riemann sphere, the complex plane, and the unit disk. For a surface of $g>=2$, the universal cover must be the unit disk. Then the deck transformations of the cover must be biholomorphisms of the unit disk, and those are precisely the orientation-preserving isometries of the hyperbolic plane under the Poincare disk model. It is now easy to give $X$ an atlas of hyperbolic charts consisting of lifts of evenly covered neighborhoods of $X$ to the universal cover.
The key point here is that a properly discontinuous group of biholomorphisms of the unit disk is exactly the same thing as a properly discontinuous group of orientation-preserving isometries of the hyperbolic plane, and thus the quotients are the same.