Quotient of the hyperbolic plane with respect to commutator group of $\pi_1(\Sigma_g)$ Let $\Sigma_g$ be a Riemann surface of genus $g\geq 2$ and $G=\pi_1(\Sigma_g)$.
Let $\pi\colon \mathbb{H}\to \Sigma_g$ be the universal covering map. What kind of surface is $\mathbb{H}/[G,G]$? 
Moreover, what is $[G,G]$; e.g. if $g=2$?
 A: $\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\RR}{\mathbb{R}}$Let $S = \Sigma_2$ be the genus two surface.  In this case, $\ZZ^4$ is the deck group of the desired covering.  Consider $\ZZ^4$ inside of $\RR^4$ and add to these points the usual edges labelled $a, b, c, d$ parallel to the four coordinate axes.  This gives a Cayley graph for $\ZZ^4$. 
Next, starting at every vertex of the graph we attach a two-cell via the attaching map $abcdABCD$ (capital letters denote inverses).  This is possible because the boundary word describes a closed loop in the graph.  Let $S'$ be the resulting two-complex.  Every edge of $S'$ meets a pair of two-cells while every vertex meets eight two-cells.  The eight corners give the vertex a disk neighborhood in $S'$.
Thus $S'$ is a surface.  Taking the quotient by the action of $\ZZ^4$ gives the original surface $S$.  By the Galois correspondence, $S'$ is the desired covering space.  Note that $S'$ is quasi-isometric to $\ZZ^4$ so it is one-ended.  The loops $abAABa$ and $cdCCDc$, based at the origin, meet in exactly one point.  Thus $S'$ has genus, and so has infinite genus. 
This construction works in any genus.  When $g = 1$ the construction produces the universal cover. 
A: Let me start by interpreting the question "What kind of surface is $S$?" in the case of a general connected oriented topological surface (without boundary). (I am considering only oriented surfaces just for simplicity of discussion.) 
If $S$ had finite complexity, i.e., would be homeomorphic to the interior of a compact oriented surface, you probably would be satisfied by the answer of the type "$S$ is has $n$ ends and genus $g$", since this provides a complete set of topological invariants. Surfaces of infinite complexity are also classified by a certain set of invariants:


*

*Its set of ends (regarded as a topological space).

*Its genus. 

*Its set of ends with positive genus.  
You can find more details and references in this MO post. 
If you look closely at the surface you are interested in, $H^2/[G,G]$, you realize that its invariants are:


*

*The surface is 1-ended (simply because the abelian group $G/[G,G]$ is 1-ended). 

*It has infinite genus (this is easy to see and is explained in Sam's answer). 

*In particular, its only end has positive genus.  
To summarize: Your surface is the unique connected oriented topological surface of infinite genus and one end. If you are looking for a different answer, you should clarify what does your question really mean.   
A: Consider the Abel-Jacobi map $\mu : \Sigma_g \to J(\Sigma_g )=\mathbb{C}^g/\Lambda$.
Then take the lift $\widetilde{\mu}: \mathbb{H} \to \mathbb{C}^g$ from the universal covering $\mathbb{H}$ of $\Sigma_g$. It seems to me that the image $X := \widetilde{\mu}(\mathbb{H})$ is the surface $\mathbb{H}/[G,G]$.
