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 Mayer-Vietoris sequence is defined as the long exact sequence
$$H_*(U \cap V) \to H_*(U) \oplus H_*(V) \to H_*(X)$$
where the first map is defined by $(i_*, j_*)$ and the second map is defined by $k_* - l_*$ where $i, j$ are the inclusion maps of $U \cap V$ inside $U$ and $V$ and $k, l$ are the inclusions of $U$ and $V$ inside $X$.
If $X$ is the Klein bottle, $\{U, V\}$ cover of it by the two mobius strips, then $U \cap V$ is homotopy equivalent to the boundary of the two strips. I have to thus compute the maps $i_*, j_*$ induced on homology where $i, j$ are inclusions of the boundary map of the mobius strips into the Klein bottle.
That's precisely where we need the wrapping description. If $M$ is a moebius strip then the inclusion map $\partial M \hookrightarrow M$ could be understood as follows: deformation retract $M$ to it's core circle, which generates $H_1(M)$. Under this deformation retract $\partial M$ maps to twice the core circle. So $i_* : H_1(\partial M) \to H_1(M)$ is precisely the multiplication by $2$ map.
Thus, $i_*$ is multiplication by $2$ whereas $j_*$ is multiplication by $-2$ due to orientation issues. Tat tells you $\alpha$ sends $1$ to $(-2, 2)$ once you identify $H_1(U \cap V)$ with $\Bbb Z$ and $H_1(U) \oplus H_1(V)$ with $\Bbb Z \oplus \Bbb Z$.
Best Answer
This is a $\Delta$-complex with one 0-simplex, three 1-simplicies, and two 2-simplicies.
Note that you have more flexibility in gluing when using a CW structure. In particular, you can construct the Klein bottle with one 0-cell, two 1-cells, and one 2-cell (think of just using a sheet of paper and gluing the edges as instructed in the fundamental polygon, so you don't have to worry about the edge "$c$" in particular).
$\pi_1(K) \cong \langle[a],[b] : [a][b] = [b][a]^{-1} \rangle$. This can be done via Van Kampen and decomposing the polygon (of $K$, without the edge $c$) into two open sets $A$ and $B$ where you can take $A$ to be the union of all edges thickened a bit, and $B$ to be the inside of the polygon with a bit of overlap with $A$ so that $A \cap B$ looks like a frame.
Look at the above picture and do the computation. You should get, with coefficients in $\mathbb{Z}$, $H_0(K) = \mathbb{Z}$, $H_1(K) = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$, and $H_2(K) = 0$ (and higher groups vanish too). A sanity check is: $K$ is connected, a surface, and nonorientable, forcing $H_0$ to be $\mathbb{Z}$ and $H_2$ to be $0$. We calculated the fundamental group and Hurewicz tells us $H_1$ is the abelianization of $\pi_1$.
The complex looks like (due to the above CW structure) $$0 \to \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z} \to 0$$ The only nonzero map is the one from $\mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$. This, from the picture above, is given by (WLOG) multiplication by $2$ in the first coordinate and $0$ in the second, assuming the first $\mathbb{Z}$ is generated by $a$ and the second by $b$. Now, that you have all the maps, you will get the same answer that you got in $(3)$.
See Hatcher (for example Theorem 2.35).
See Hurewicz theorem or Hatcher 2.A. The basic intuition you should go in with is that $f: I \to X$ can be viewed as both a path and a singular 1-simplex.