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 $1 \mapsto (2,0)$.
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
See here: the coefficients are the degree of the attaching map. Yes, your last sentence is correct: if $e_0 \mapsto v_0 -v_1$ then $e_1 \mapsto v_1 - v_0$.