The answer is simple: any complex structure arises in this way.
Indeed, any compact Riemann surface $X$ admits a holomorphic cover $\phi \colon X \to \mathbb{P}^1$.
This is straightforward in genus $0$ and $1$.
If the genus is at least $2$, then the linear system $|3K_X|$ is very ample, so it gives an embedding $\gamma \colon X \to \mathbb{P}^N$ and the map $\phi$ is obtained by composing $\gamma$ with a suitable projection.
We are using here the fact that every compact Riemann surface is a smooth complex projective curve; this follows from the existence of a meromorphic function on it, which is a non-trivial fact (see Andy Putman's comment).
EDIT. We can always chose $\phi$ so that the local monodromy is general, that is it is given by a single transposition around each branch point. In fact, projecting $X \subset \mathbb{P}^N$ from a general subspace of the right codimension, we obtain a plane curve $X' \subset \mathbb{P}^2$ whose singularities are at worst ordinary double points. Moreover, $X'$ has at most a finite number of bitangent lines; taking the projection $\pi_p \colon X' \to \mathbb{P}^1$ from a point $p$ not contained in any of the bitangent lines and composing with the map $X \to X'$ we obtain a finite cover $\phi \colon X \to \mathbb{P}^1$ whith the desired property.
Q1. There are two DIFFERENT notions of Riemann surface in the literature.
a) One-dimensional complex analytic manifold (coming from the book of Weyl).
b) Riemann surface "spread over the plane (or over the Riemann sphere)". Your second
definition, the set of germs with an appropriate topology on it, formalizes this second notion.
Older books seem to understand Riemann surfaces in the sense of the second definition.
Sometimes a) was called an "abstract Riemann surface" in these books.
For most mathematicians with
modern training the "Riemann surface of log z" and the
"Riemann surface of arccos z" are meaningless expressions because these
are the same as the plane, in the sense of definition a).
The formal relation between a) and b) is the following.
"A Riemann surface spread over the plane" is a pair (S,f), where S is an abstract
Riemann surface and f is a holomorphic function from S to C.
(If f is meromorphic, we have a Riemann surface spread over the sphere.)
Here is another way to say this. Let S be a Riemann surface in the sense a).
It has a set of charts $\phi_j: U_j\to D_j$ from the elements of an open covering U
to discs D in the plane. The correspoddence maps $\phi_k\circ\phi_j^{-1}$
on $D_j\cap D_k$ must be conformal.
Now let us require that these correspondence maps be IDENTITY maps of
$D_j\cap D_k$. Then we obtain notion b). This is an additional structure on
a Riemann surface in the sense a) which is sometimes called a flat structure.
If you look carefully (say, on the example of arccos) you will see that the two
definitions of a Riemann surface in the sense b) that I gave are not exactly equivalent.
More about this in my survey "Geometric theory of meromorphic functions", and
in the preprint of Biswas and Perez Marco, Log Riemann Surfaces.
Best Answer
A counterexample to Q3 is provided by the genus 2 compact Riemann surface $X$ of $y^2=x^5-1$. Indeed, the order 10 cyclic group $C_{10}$ acts on $X$ (by changing sign of $y$ and multiplying $x$ by $5$th roots of unity). It is known that the jacobian of $X$ has endomorphism ring $Z[\zeta_5]$ - the $5$th cyclotomic ring of integers and any finite multiplicative subgroup of $Z[\zeta_5]$ is a subgroup of $\mu_{10}\cong C_{10}$. This implies that $Aut(X)=C_{10}$. On the other hand, the dihedral group $D_{10}$ of order $10$ has the same Sylow subgroups as $C_{10}$ but is not isomorphic to it. In other words, there is no faithful action of $D_{10}$ on $X$ while its Sylow subgroups $C_5$ and $C_2$ act faithfully on $X$.
If $Y$ is a compact Riemann surface of genus $g$ and its jacobian $J$ has no nontrivial automorphisms (i.e., $End(J)$ is the ring of integers $Z$) then either $Y$ is non-hyperelliptic and $Aut(Y)=\{1\}$ or $Y$ is hyperelliptic and $Aut(Y)=C_2$. For example, if $g>1$ and $Y_g$ is the hyperelliptic Riemann surface $y^2=x^{2g+1}-x-1$ then its jacobian $J_g$ has no nontrivial endomorphisms (Math. Research Letters 7 (2000), 123--132) and therefore $Aut(Y_g)=C_2$. If $p$ is an odd prime then for each integer $n \ge 5$ the automorphism group of the compact Riemann surface $y^p=x^n-x-1$ is the cyclic group $C_{p}$. Indeed, the endomorphism ring of the jacobian is the $p$th cyclotomic ring $Z[\zeta_p]$ (Math. Proc. Cambridge Philos. Soc. 136 (2004), 257--267) and one may easily check, using the differentials of the first kind that the curve is non-hyperelliptic.
Using Del Pezzo surfaces of degree 2, one may construct non-hyperelliptic genus 3 curves $Y$, whose jacobian has no nontrivial endomorphisms (AMS Translations Series 2, vol. 218 (2006), 67--75; MR2279305, 2007k:14060) and therefore $Aut(Y)=\{1\}$. For the genus 4 case see a paper of Anthony Várilly-Alvaradoa and David Zywina (LMS Journal of Computation and Mathematics (2009), 12: 144-165); their approach makes use of Del Pezzo surfaces of degree 1 (see also Math. Ann. 340 (2008), 407--435).