Every Riemann Surface has a countable basis for its topology

This theorem is proven in the book by Forster (as indeed listed in the references of your book). There is no assumption of countability, as far as I can tell. I'll give you the relevant definitions (and the theorem), verbatim from the book:

Definition 1: An $n$-dimensional manifold is a Hausdorff topological space such that every point in it has an open neighborhood homeomorphic to an open subset of $\Bbb R^n$.

Definition 2: A Riemann surface is a pair $(X,\Sigma)$ where $X$ is a connected 2-manifold and $\Sigma$ is a complex structure on $X$.

Theorem (Rado): Every Riemann surface has a countable topology.

The theorem seems to follow from the existence of solutions to the Dirichlet problem (discussed in section 22 of the book), with the proof of the theorem residing in section 23.

In particular, countability is not assumed anywhere. As you mention, countability is an oft-used assumption/axiom in differential geometry, but in the theory of Riemann surfaces it appears to be redundant.

In his book (Lectures on Riemann surfaces), Forster define a Riemann surface as an Hausdorff space with holomorphic atlas.

This is a theorem of Rado, proved in the same book (page 186), that these conditions implies that a Riemann surface has a countable basis.

I have the same feeling as you though : in the definition of a manifold, the countable basis is usually required since it is a crucial tool for partitions of unity.