I am interested in calculating of $H_2(G,\mathbb{Z})$, where $G$ is a group given by presentation with two relators $\langle a,b| r_1 = r_2 = 1\rangle$.
Moreover, I am interested in such presentations, that $H_2(G,\mathbb{Z}) = \mathbb{Z}^2$, it would be great to learn how to construct a lot of such examples.
If $G$ is a perfect group (i.e. $G=[G,G]$, i.e. $G_{ab}=0$) which admits a presentation with two generators and two relations, then $H_2G=0$.
More generally, if $G$ is a group which admits a presentation with $n$ generators and $m$ relations then $H_2G$ can be generated by $m-n+\text{rk}_\mathbb{Z}(G_{ab})$ elements. This appears as Exercise II.5.5(b) in Ken Brown's group cohomology book. Here is my old solution:
With the presentation $G=\langle s_1,\cdots,s_n\,|\,r_1,\cdots,r_m\rangle$ we associate the $2$-complex $\;Y=(\bigvee_s S^1)\cup_{r_1}e^2\cdots\cup_{r_m}e^2\;$ so that $\pi_1Y\cong G$. By computing the Euler characteristic $\chi(Y)$ two different ways we obtain the equation $\sum(-1)^i\text{rk}_\mathbb{Z}(H_iY)=\sum(-1)^ic_i$, where $c_i$ is the number of $i$-cells. Then $1-\text{rk}_\mathbb{Z}(G_{ab})+\text{rk}_\mathbb{Z}(H_2Y)=1-n+m$ and so $\text{rk}_\mathbb{Z}(H_2Y)=m-n+r$, where $r=\text{rk}_\mathbb{Z}(G_{ab})=\dim_\mathbb{Q}(\mathbb{Q}\otimes G_{ab})$. Now $H_2Y=\text{ker}(\partial_2)$ is a free abelian group (subgroup of cellular 2-chain group), and by applying Theorem II.5.2[Brown] we get a surjection $H_2Y\rightarrow H_2G$ (from the exact sequence in that theorem -- this theorem is a play on the Hurewicz theorem). Thus $H_2G$ $\textit{can}$ be generated by $m-n+r$ elements. Now in my special scenario, $m-n+r=2-2+0=0$.
First, I am very skeptical that one can, in general, calculate $H_2(G, \mathbb{Z})$ using just a presentation $\langle a,b \, \vert \, r_1=r_2 =1 \rangle$. It is generally known that finite presentations for groups are inadequate to solve many kinds of decision problems. You will probably need to know more about $G$ to calculate $H_2(G, \mathbb{Z})$ using either the Hopf formula or spectral sequences.
Second, if you do know of one example of a group $G$ which has a presentation
$\langle a,b \, \vert \, r_1=r_2 =1 \rangle$ and such that
$H_2(G, \mathbb{Z})$, is $\mathbb{Z}\times \mathbb{Z}$, then it is not difficult to construct lots of other
presentations for $G$. Let $F$ be the free group on $\{a,b\}$. You can use a sequence of regular
elementary Nielsen transformations (See Lyndon and Schupp, Chapter I, Section 2) to find another
pair of words, $\{A,B\}$ that also freely generate $F$. Write $R_i, i=1,2$ for the word obtained from $r_i$
by replacing every $a$ by $A$ and every $b$ by $B$ (and doing the same replacement for inverses,
of course). Then $\langle A,B \, \vert \, R_1=R_2 =1 \rangle$ is also a presentation for $G$. Then
$a$ can be written as a word on $A$ and $B$ and $b$ can be written as a word on $A$ and $B$, so we can
write $r_1$ and $r_2$ as words on $A$ and $B$ and obtain a presentation
$\langle A,B \, \vert \, r_1=r_2 =1 \rangle$ for $G$. Dually, we can rewrite $R_1$ and $R_2$ as words on
$a$ and $b$ and obtain a presentation $\langle a,b \, \vert \, R_1=R_2 =1 \rangle$. Similarly,
we could obtain a new presentation for $G$ by swapping $a$ with $b$ in $r_1$ and $r_2$. Also we
can replace $r_1$ and
$r_2$ by conjugates of these.
Finally, I will mention one example which many might consider to be obvious. Let $G$ be the group on 2 generators which is free in the variety of groups having nilpotence class 2. Then $G$ has numerous presentations of the form $\langle a,b \, \vert \, r_1=r_2 =1 \rangle$ and the Schur multiplier, $H_2(G, \mathbb{Z})$, is $\mathbb{Z}\times \mathbb{Z}$.
Write $\{\gamma_k(H)\}$ for the lower central series of any group $H$: i,e. $\gamma_1(H) = H$ and $\gamma_{k+1}(H) = [\gamma_k(H),H]$. Let $F$ be the free group on two generators, $a$ and $b$. Let $G$, as above, be the group on 2 generators which is free in the variety of groups having nilpotence class 2. Then $G = F/\gamma_3(F)$ and one presentation for $G$ is $\langle a,b \, \vert \, [b,a,a]=[b,a,b]=1\rangle$. Here, the smallest normal subgroup, $R$, containing $[b,a,a]$ and $[b,a,b]$ is $\gamma_3(F)$. It is not too difficult to see that $F/\gamma_4(F)$ is the covering group for $G$ and that $\gamma_3(F)/\gamma_4(F)$ is the Schur multiplier. It is also easy to verify that $\gamma_3(F)/\gamma_4(F)$ is free abelian with generators $[b,a,a]\gamma_4(F)$ and $[b,a,b]\gamma_4(F)$.
Later (November 16, 2015): For another example for $G$, let $G$ be the group having presentation $\langle a,b \, \vert \, [b,a,a] = [b,a,b,b]=1 \rangle$.
