The simplest case, where $g=0, n=3$ and $g = n = 1$ yield isomorphic groups (free of rank 2), can be written explicitly using a little $2 \times 2$ matrix calculation. We choose a fundamental domain in the upper half-plane made out of two vertical lines with real parts $-1$ and $1$, and two semicircles whose diameters are the real intervals $[-1,0]$ and $[0,1]$. Since the boundaries are geodesics, it suffices to find Möbius transformations that transform the endpoints appropriately.
For $g=0, n=3$, we choose generators $\begin{pmatrix}1& 2 \\ 0 & 1 \end{pmatrix}$ to glue the vertical lines together, and $\begin{pmatrix}1& 0 \\ 2 & 1 \end{pmatrix}$ to glue the semi-circles. If you like modular curves, this quotient is called $Y(2)$, and classifies isomorphism classes of elliptic curves equipped with an ordered list of all 2-torsion points.
For $g=1, n=1$, we choose $\begin{pmatrix}1& 1 \\ 1 & 2 \end{pmatrix}$ to glue the left vertical line to the right semicircle, and $\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ to glue the left semicircle to the right vertical line. This quotient is a leaky torus.
The above results form a special case of a general phenomenon (mentioned by Sam Nead), where the loops around punctures give unipotent (aka parabolic) generators of $\pi_1$, and handles give a pair of hyperbolic generators.
I don't have a good answer concerning the set of lifts of the boundary points - it will always be a disjoint union of $n$ orbits under the transformation group, but it can vary widely, since the transformation group has a continuous family of representations in $PSL_2(\mathbb{R}$.