Most topologists would be happy just drawing the diagram you've drawn (though the topologists I know prefer to draw on apples), but if you want to do it explicitly then you can as well.
As you know, the torus $S^1\times S^1$ is homeomorphic to $[0,1]\times [0,1]/\equiv$, where $\equiv$ identifies the edges of the square by $(x,0)\equiv(x,1)$ and $(0,y)\equiv(1,y)$. We also define the Klein bottle to be $K=[0,1]\times [0,1]/\sim$, where $\sim$ identifies the edges of the square by $(x,0)\sim(x,1)$ and $(0,y)\sim(1,1-y)$.
For the torus, we have an explicit continuous surjection
$$
\pi:[0,1]\times[0,1]\to S^1\times S^1: (x,y)\mapsto\left(e^{i\pi x},e^{i\pi y}\right)
$$
using the standard identification of $S^1$ with the unit circle in the complex plane (more a notational convenience than anything else). Note that we now have:
$$
(x_1,y_1)\equiv(x_2,y_2)\Longleftrightarrow \pi(x_1,y_1)=\pi(x_2,y_2)
$$
In other words, $\pi$ induces a well-defined homeomorphism $([0,1]\times[0,1]/\equiv)\to S^1\times S^1$.
The next step is to interpret your diagram as a map $[0,1]^2\to[0,1]^2$. This map is then going to induce the two-sheeted covering we want. Explicitly, we have:
$$
\phi:[0,1]\times[0,1]\to[0,1]\times[0,1]: (x,y)\mapsto
\begin{cases}
(2x,y) &\mbox{if } x\le\frac12 \\
(2x-1,1-y) & \mbox{if } x\ge\frac12.
\end{cases}
$$
Composing this map $\phi$ with the projection $\pi_\sim:[0,1]\times[0,1]\to K$, we get a map $\pi_\sim\circ\phi : [0,1]\times[0,1] \to K$.
We claim that this map $\pi_\sim\circ\phi$ induces a two-to-one covering map
$$\psi : S^1 \times S^1 \,\,\, = \,\,\, [0,1]\times[0,1]/\equiv \,\,\,\to\,\,\,[0,1] \times [0,1] / \sim \,\,\,= \,\,\,K
$$
Proving that $\psi$ is two-to-one means checking
$$
|(\psi^{-1}(\{q\})/\equiv)|=2
$$
for each $q \in K$. And to prove that $\psi$ is a covering map it suffices to check that $\psi$ is a local homeomorphism at $p \in S^1 \times S^1$ (ordinarily this is not enough for checking that something is a covering map, but it suffices when the domain and range are compact manifolds). So one has to check something for the points in $[0,1] \times [0,1]$ that form the equivalence class of the relation $\equiv$ corresponding to $p$: the four corner points; or a pair of opposite side points; or an interior point. Namely one must find neighborhoods of those points which, when fitted together under $\equiv$, form an open neighborhood of $p$ that maps homeomorphically onto an open neighborhood of $q=\psi(p)$. Checking these things is the real content of the proof, and I'll leave them as exercises. It's basically what your diagram is telling you.
Now we have a double-cover by $[0,1]\times[0,1]/\equiv$ of $K$. We already remarked that there is a homeomorphism between $S^1\times S^1$ and $[0,1]\times[0,1]/\equiv$; putting these together gives us a double cover of $K$ by $S^1\times S^1$.
I should stress - there's very little content in any of this, and it really is just a way of making your diagram 'rigorous' in some sense. It's good to work thorugh a few examples like this one explicitly, but you'd be bananas to try and be completely rigorous all the time in topology.
The third covering space can also be viewed by "splitting" the Klein bottle $K$, if you are careful about what that means. In each case there is a simple closed curve $c \subset K$ such that if you first let $L_c = K \setminus c$ be the compact surface-with-boundary obtained by cutting $K$ open along $c$, then the covering space is obtained by taking the disjoint union of two copies $L_1,L_2$ of $L_c$, by gluing the boundaries $\partial L_1$, $\partial L_2$ using an appropriately chosen homeomorphism $\partial L_1 \mapsto \partial L_2$.
Careful choice of $c$ for each of the three homomorphisms is what distinguishes the three cases, and the idea is to choose $c$ to be a simple closed curve in $K$ representing an appropriate element $g_c \in \pi_1 K$, chosen by two criteria: $a \mapsto 0$ if and only if $a$ is represented by a closed curve in $L_c$; and $b \mapsto 0$ if and only if $b$ is represented by a closed curve in $L_c$. Here are the three cases, with a description of $L_c$ (I worked this out with some pictures, so I do THINK it's right):
- For $a \mapsto 0$, $b \mapsto 1$, choose $g_c=a$. $L_c$ is an annulus.
- For $a \mapsto 1$, $b \mapsto 0$, choose $g_c=b$. $L_c$ is a Möbius band.
- For $a \mapsto 1$, $b \mapsto 1$, choose $g_c=aba^{-1}$. $L_c$ is a Möbius band.
Best Answer
One subgroup of $\langle a, b\ |\ abab^{-1} = 1 \rangle $ that is isomorphic to $\mathbb{Z}\times \mathbb{Z}$ is the subgroup generated by $a$ and $b^2$. Indeed these elements commute: $ab^2 = ba^{-1}b = b^2 a$, and as this is a finite-index sub-group of a surface group it is also a surface group, so in particular it only has one relation. In general you could take $a^m$ and $b^{2n}$.
In an answer to a related question (Is there a non-trivial covering of the Klein bottle by the Klein bottle?) I gave a few families of subgroups and determined the total spaces of the corresponding cover space in those cases, maybe you will find it useful.