Consider a map $f\colon S^1 \to S^1 \vee S^1$ that maps upper half of a circle to the first summand in orientation preserving way, and lower half to the second circle in orientation reversing way. Also denote $i_1$ and $i_2$ inclusions of $S^1$ to $S^1 \vee S^1$ as two summands. Composition of $f$ with folding map $p\colon S^1\vee S^1 \to S^1$ (both summands in orientation-preserving way) is evidently homotopic to a constant map. Looking on $H_1$ we have $\mathbb{Z} \xrightarrow{f_*} \mathbb{Z}\oplus \mathbb{Z} \xrightarrow{p_*} \mathbb{Z}$.
$p_*$ is addition: $S^1 \xrightarrow{i_1} S^1\vee S^1 \xrightarrow{p} S^1$ is identity (similary for $i_2$). $i_1$ induces $\mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z}$, to make sure it is an inclusion of the first summand we may use naturality of exact sequences for a map $(S^1, *) \xrightarrow{i_1} (S^1\vee S^1, \mathrm{im}\, i_2)$. Now we know $p_*(1, 0) = 1$ and $p_*(0, 1) = 1$, by linearity $p_*$ is addition.
Thus $f_*$ maps $1\in H_1(S^1)\cong \mathbb{Z}$ to some pair $(x, -x)\in H_1(S^1\vee S^1)$. To find out what $x$ is, consider the composition $S^1 \xrightarrow{f} S^1 \vee S^1 \to S^1$ where the second map collapses second summand to a point (it induces projection on a second coordinate in $\mathbb{Z}\oplus \mathbb{Z} \to \mathbb{Z}$ too, e.g. by naturality of exact sequences of pairs). By construction this map is homotopic to identity. Similary for the second summand in $S^1\vee S^1$ we get that orientation-reversing map induces $-1$ on homology.
That's quite cumbersome to read, but actually very straightforward.
UPDATE
As mentioned in comments, here is proof reversing orientation induces identity on $H_0$. $* \to S^1$ and $*\to S^1 \to S^1$ (the second map is reversal of orientation) are equal as maps from a point, thus induce the same map on $H_0$. If we know that the inclusion of a base point in $S^1$ induces non-zero map on $H_0$, we're done, since automorphism of $H_0(S^1) \cong \mathbb{Z}$ fixing a non-zero element must be identity. To see this, use the naturality of long exact sequences of pairs for a map $(S^1, *) \to (*, *)$. The first commutative square (with non-relative $H_0$s) consists of three identity mapping of a point and a map induced from inclusion of a base point, and cannot commute unless $H_0(*) \to H_0(S^1)$ is an inclusion.
(Reply to a comment: nullhomotopic map induces zero on $H_1$ since it factors as map $X \to * \to Y$)
Also it seems to me that $X \hookrightarrow X \vee Y$ induces inclusion to the first summand in homology not due to functoriality properties of long exact sequences but this is just the way we identify $H_i(X\vee Y)$ with $H_i(X)\oplus H_i(Y)$. That is, we look at the pair $(X \vee Y, Y)$, write long exact sequence, then apply excision to get rid of relative homology and get $H_i(Y)$ instead. After that, we note that a lot of arrows in long exact sequence admit splitting and define the isomorphism $H_i(X \vee Y) \to H_i(X) \oplus H_i(Y)$ the way inclusion of $X$ in a wedge sum is inclusion on first summand. Thus some arguments by naturality above are not needed.
To finish off the comments...
Everything is good up through the cellular chain complex, given below, which is only nonzero in degrees zero through two:
$\cdots \to 0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z} \to 0$
Then you had the right answer, and a comment I made caused a bit of confusion which led me to point out something important: the cellular chain complex is (in general) not exact, and computing its homology gives you the homology of your space. So everything you added in your edit starting with "The exact sequence of homology groups is..." is not how you'll want to finish it off (my comment had meant something more like, "finish it off by adding a sentence for how you can tell $H_2 \cong 0$ when you're reading off the homology of your chain complex").
Edit: The OP got edited so this above paragraph is more or less irrelevant.
Anyways, now let's compute the homology of $\mathbb{R}P^2$, which amounts to the homology of the above chain complex. Label the maps $\delta_1$ and $\delta_2$. Then $H_0(\mathbb{R}P^2) \cong \mathbb{Z}/\text{im }\delta_1 \cong \mathbb{Z}/0 \cong \mathbb{Z}$.
$H_1(\mathbb{R}P^2) \cong \ker\delta_1/\text{im }\delta_2$. Since $\delta_1$ is zero its kernel is all of $\mathbb{Z}$, and since $\delta_2$ is multiplication by $2$ its image is $2\mathbb{Z}$, so $H_1(\mathbb{R}P^2) \cong \mathbb{Z}/2\mathbb{Z}$.
$H_2(\mathbb{R}P^2) \cong \ker\delta_2/\text{im }0 \cong \ker \delta_2$. Since $\delta_2$ is multiplication by $2$, it is injective ($\mathbb{Z}$ is an integral domain), so $\ker \delta_2 = 0$. Thus $H_2(\mathbb{R}P^2) \cong 0$.
Since the cellular chain complex consists of zeros above degree $2$, $H_i(\mathbb{R}P^2) \cong 0$ for $i > 2$ as well.
Best Answer
For the 1-cell attaching maps the degree is easy: your attaching maps are constant because they are the maps $S^0 \to X^0 = \{x_0\}$ sending both endpoints of your $1$-cells to the $0$-cell. Thus both $\text{deg}f_{a,x_0}$ and $\text{deg}f_{b, x_0}$ are zero and thus $\delta_1 = 0$. It's worth noting that in general, if your $1$-skeleton ends up a wedge of circles the cellular boundary $\delta_1$ will always be zero, for this reason.
Now, I'll call the $2$-cell $e$, instead of $ab$ as you've done, so that when I refer to the $2$-cell it's a bit clearer (its attaching map will involve $a$'s and $b$'s so I don't want any confusion there).
$\delta_2$ also has a quick strategy for computation. The first step is usually to describe the attaching map for the $2$-cell in terms of the $1$-cells; in this case, we can say that the attaching map is $baba^{-1}$ (read off the edges in your polygon), corresponding to $\gamma: S^1 \to X^1 = S^1 \vee S^1$ that on the first quarter-circle of the domain traces the 1-cell $b$, on the second quarter-circle traces $a$, etc.
Now that we've done this, realize that $f_{e,a}$ is the attaching map restricted to $a$, so basically we're deleting $b$ from the formula for $\gamma$; this is the interpretation of the map $f_{\alpha,\beta}$ that you've described. This means that $\text{deg}f_{e,a}$ is the degree of the map described by $aa^{-1}$, which is constant. Similarly, $\text{deg}f_{e,b}$ is the degree of the map represented by $b^2$, which has degree $2$.
Therefore $\delta_1: \mathbb{Z} \oplus \mathbb{Z} \to \mathbb{Z}$ is the $0$ map and $\delta_2: \mathbb{Z} \to \mathbb{Z} \oplus \mathbb{Z}$ sends $1 \mapsto (2,0)$.