For the new question an answer is given by arithmetic Fuchsian groups. For example it is well-known that the reflection group associated with the regular right-angled pentagon in $\mathbb H^2$ contains every surface group as it contains the fundamental group of the non-orientable surface of Euler characteristic -1 as an index-4 subgroup. On the other hand it is an index-10 subgroup in the (2,4,5)-triangle group $\Delta$ (by dividing the pentagon into triangles from the center with vertices on the pentagon's and on the middles of its edges). The latter is known to be arithmetic by a result of Takeuchi (see for example item 6 in the table in section 13.3 of Maclachlan--Reid https://zbmath.org/?q=an%3A1025.57001). Its trace field is $\mathbb Q(\sqrt 5)$, which means that $\Delta$ will be realisable as a subgroup of $\mathrm{PSL}_2$ over a quadratic extension $F$ of $\mathbb Q(\sqrt 5)$, and contained in the ring of integers $\mathbb Z_F$. There are many choices for such an extension, for example one can take $F = \mathbb Q(\sqrt 2, \sqrt 5)$ since the quaternion algebra ramifies only at primes dividing 2. So in principle you get an embedding of $\Delta$, and hence of any surface group, into $\mathrm{PSL}_2(\mathbb Z_F)$. You can ask Sage to compute an integral basis for the latter to get a complete answer to your question.
Another possibility to avoid the arithmetic machinery would be to use hyperbolic geometry to compute generators for the triangle group directly and see where they lie, this has probably been done but I don't know a reference (and I'm not sure how to choose the walls as to get something which would lie where it should, i.e. $\mathbb Z_F$).
This may be not optimal, maybe all surface groups embed into $\mathrm{PSL}_2$ over a real quadratic field. For example the (2, 4, 6)-triangle group is arithmetic with trace field $\mathbb Q$, it embeds into $\mathrm{PSL}_2(\mathbb Z[\sqrt 2])$ if I'm not mistaken (see the table in Maclachlan--Reid) so its surface subgroups do as well. However I don't know which genera will be realised this way.