To rephrase Alex Bartel's answer geometrically: the quotient of the hyperbolic plane by $\Gamma(2)$ is a 3-punctured sphere, whose homology is free abelian of rank 2. Any space has a universal abelian cover, whose group of deck transformations is its homology. The universal abelian cover is easier to see for the punctured torus, which has the same fundamental group --- it's just the plane minus holes for the punctures. The homology of this cover is coordinatized by linking number with the holes.
To picture the universal abelian cover of the 3-punctured sphere: enlarge the punctures to make boundary components, forming a pair of pants. Cut the pair of pants along 3 seams into two hexagons. Use a template consisting of the plane divided into equilateral triangles with an alternating coloring, say red and blue; color the hexagons red and blue. Above each red triangle, put a red hexagon above it in $\mathbb R^3$ so that three ofthe hexagon's edges project to the midpoints of edges of the red triangles, and the other three connect the top of one vertical edge to the bottom of the counterclockwise vertical edge. Arrange the blue hexagons in the same way, over the blue triangles. Half of the edges of hexagons string together to form lines that weave in and out, in a hexagonal weave: these cover the three boundary components of the pair of pants.