The short answer is that your presentation is isomorphic to the (slightly more) "standard" presentation - so you computed a correct answer.
More detail -
According to Wolfram and using $K$ to denote the Klein bottle, we have $\pi_1(K) \cong \langle c,d \rangle/cdc^{-1}d$ whereas you have (according to Aaron Mazel-Gee's comment) $\pi_1(K)\cong \langle a,b\rangle /abab^{-1}$.
The question is, then, are these isomorphic?
Well, we have $cdc^{-1}d = e$ so, taking the inverse of both sides gives $d^{-1}cd^{-1}c^{-1} = e$. But this has the same form as the relation between $a$ and $b$, so now the isomorphism is clear: we map $a$ to $d^{-1}$ and $b$ to $c$.
That is, define $f:\langle a,b \rangle /abab^{-1}\rightarrow \langle c,d\rangle/cdc^{-1}d$ by $f(a) = d^{-1}$ and $f(b) = c$.
I claim this is well defined, for \begin{align*} f(abab^{-1}) &= f(a)f(b)f(a)f(b)^{-1} \\\ &= d^{-1}cd^{-1}c^{-1} \\\ &= (cdc^{-1}d)^{-1} \end{align*} as it should. (Technically, I'm defining $f$ on $\langle a,b\rangle$ and proving it descends to the quotient.)
Finally, rather than show this is 1-1 and onto, instead, show that $g:\langle c,d\rangle/ cdc^{-1}d\rightarrow \langle a,b\rangle/ abab^{-1}$ defined by $g(c) = b$ and $g(d) = a^{-1}$ is well defined, and the inverse to $f$, so $f$ is the desired isomorphism.
The concept you're probably looking for is the deck transformation group of a covering space. In this case, your group $G$ happens to be an invariant composite of the covering $\phi$, that is, for all $\rho\in G$, the composition $\phi\circ\rho=\phi$. It follows that, as $\mathbb{R}^2/G$ is the full orbit space, it must be that $G$ is the full deck transformation group for the covering $\phi$.
Now, $\mathbb{R}^2$ is a simply connected topological space (in fact it's contractible) and so it is not just a cover but a universal cover of the Klein bottle. Universal covers have the nice property of always being regular coverings (the induced action of the deck transformation group on the fiber of the cover is both free and transitive) and so the deck transformation group $G$ is isomorphic to the fundamental group $\pi_1(\mathbb{R}^2/G)$ of the Klein bottle.
The justification of the last claim is easiest to describe using the theory of principle $G$-bundles although this might be a bit out of the scope of your current course so feel free to ignore this if you get lost (I believe Hatcher gives a very good elementary explanation of the isomorphism without reference to bundles, so feel free to check his textbook). For completeness though, we'll show that $G\cong\pi_1(\mathbb{R}^2/G)$. Let $F$ be the fiber of the map $\phi$. The principle $G$-bundle $\phi$ induces an exact sequence in homotopy $$\pi_1(\mathbb{R}^2)\stackrel{\phi_*}{\longrightarrow}\pi_1(\mathbb{R}^2/G)\longrightarrow\pi_0(F)\longrightarrow\pi_0(\mathbb{R}^2).$$ Now $\pi_1(\mathbb{R}^2)$ is trivial, and $\mathbb{R}^2$ is path connected so $\pi_0(\mathbb{R}^2)$ is trivial. We know that $\pi_0(F)$ is actually a group and is isomorphic to $G$ because $\phi$ is a principle $G$-bundle. It follows that we have the short exact sequence $$0\stackrel{\phi_*}{\longrightarrow}\pi_1(\mathbb{R}^2/G)\longrightarrow G\longrightarrow 0$$ and so we have an induced isomorphism $\pi_1(\mathbb{R}^2/G)\stackrel{\cong}{\longrightarrow} G$.
Best Answer
Every cyclic subgroup is abelian, hence nilpotent. Every such subgroup is generated by an element of the form $x^ay^b$ with $a$ and $b$ integers.
In addition, $y^2$ commutes with $x$. Since $y^2$ is central, to determine the commutator of two elements we only need to consider the parity of the exponents of $y$ in their normal forms. Elements of the form $x^ay^{2b}$ and $x^ry^{2s}$ commute; elements of the form $x^ay^{2k+1}$ and $x^r$ commute if and only if $r=0$; and elements of the form $x^ay^{2k+1}$ and $x^ry^{2s+1}$ commute if and only if $a=r$. That gives you lots of abelian subgroups that are isomorphic to $\mathbb{Z}\times\mathbb{Z}$.
What about higher nilpotency? Suppose you have two noncommuting elements in your subgroup. It is not hard to check that $[x^ay^{2b+1},x^r] = x^{2r}$ and $[x^ay^{2b+1},x^r y^{2s+1}] = x^{2a-2r}$. But no nontrivial power of $x$ commutes with an element of the form $x^a y^{2b+1}$; and further commutators will just yield further nontrivial powers of $x$ that still do not commute with an element of the form $x^ay^{2b+1}$. So if your subgroup $H$ has an element of the form $h_1=x^a y^{2b+1}$, and $h$ is any nontrivial element of $H$ that is not a power of $h_1$, then $h_1$ and $h$ do not commute, and $h_1$ does not commute with any of $[h_1,h]$, $[h_1,h,h_1]$, $[h_1,h,h_1,h_1],\ldots,[h_1,h,h_1,\ldots,h_1]$, etc. But if $H$ is nilpotent of class $c$, then any commutator of weight $c$ would be central in $H$. Thus, $H$ cannot be nilpotent.
So the only nilpotent subgroups of $H$ are abelian, and they are given as above.