Chain complex for an easier CW structure.
Well, first it is easier to compute the homology if you use the CW structure on $S^n$ having one $n$-cell and one $0$-cell. The cellular chain complex just looks like $$\mathbb{Z}\to0\to\cdots\to0\to\mathbb{Z}$$ or if $n=1$, then $\mathbb{Z} \overset{0}\to\mathbb{Z}$.
Then if you adjoin one more cell via a degree $m$ map, by definition this gives you a cellular chain complex $$\mathbb{Z} \overset{m}\to\mathbb{Z}\to0\cdots\to0\to\mathbb{Z}.$$ Following the instructions for the cellular boundary formula, as on page 140 of Hatcher, there is really no work to do here since in this case the complement of our cell in dimension $n$ is already just a point.
Chain complex for the CW structure having two cells in each dimension.
If you are really set on using the CW structure with two cells in each dimension, it takes a little more work to find the cellular chain complex for $S^n$. Also there is some amibiguity here in that there is more than one CW structure with two cells in each dimension that produces $S^n$. Of course the chain complexes will all be quasi-isomorphic, but if you do it in an organized way then you will get $$\mathbb{Z}^2 \overset{d_n}\longrightarrow \mathbb{Z}^2 \overset{d_{n-1}}\longrightarrow \cdots \overset{d_1}\longrightarrow\mathbb{Z}^2$$
where each of the $d_i$ are just given by the matrix $\begin{pmatrix}1&-1\\1&-1\end{pmatrix}$. I will describe how to do this below. Once you see this, then attaching the $n+1$ cell via a degree $m$ map should give you
$$\mathbb{Z} \overset{\begin{pmatrix}m\\ m\end{pmatrix}}\longrightarrow\mathbb{Z}^2 \overset{d_n}\longrightarrow \mathbb{Z}^2 \overset{d_{n-1}}\longrightarrow \cdots \overset{d_1}\longrightarrow\mathbb{Z}^2.$$
Computing $d_{n+1}$ with the cellular boundary formula.
To compute that $d_{n+1}$ is as above using the cellular boundary formula, consider the two $n$-cells in $S^n$, one for each hemisphere. The coefficient of $d_{n+1}(e^{n+1})$ on the top hemisphere is obtained by crushing the lower hemisphere and checking the degree of the map $$\partial D^{n+1} \to S^n \overset{\text{crush}}\longrightarrow S^n.$$ By the local degree theorem, this only depends on the size of the preimage at a suitable point - but this is the same as for the original map since all we did was crush the lower hemisphere. So the degree of this map is still $m$. The same logic applies to the coefficient on the cell for the lower hemisphere.
Seeing that $d_i = \begin{pmatrix}1&-1\\1&-1\end{pmatrix}$.
Label the cells in dimension $k$ by $e^k$ and $f^k$, which will be the correct ordered basis for the matrices above. Here is how to define attaching maps so that all chain maps are given by the matrix above.
Consider the $k$-skeleton $S^k$ inside of the plane $\mathbb{R}^{k+1}$ in $\mathbb{R}^{k+2}$ (you should probably take $k=1$ for visualization purposes).
Above $S^k$ in $\mathbb{R}^{k+2}$ is the top hemisphere of $S^{k+1}$ which is isomorphic to $D^{k+1}$ via the projection map. The attaching map is the identity, so as above we have that $d_{k+1}(e_{k+1}) = e_k + f_k$. The lower hemisphere of $S^{k+1}$ is isomorphic to $D^{k+1}$ as well, but the orientation of the boundary is the opposite, relative to $S^k$. To see this, note that the attaching map is the same identifying $e^k$ with the upper hemisphere, reflecting through $\mathbb{R}^{k+1}$, then attaching along the identity map. So, the cellular boundary formula tells you $d_{k+1}(f_{k+1}) = -e_k - f_k$.
For this post, I am thinking of $\mathbb{R}P^n$ modeled by the northern hemisphere of $S^n$ with antipodal equatorial points identified. Also, I am just going to write $f$ instead of $f_n$.
Let $N,S\subseteq S^n$ denote the (closed) northern and southern hemispheres. Let $n\in N$, $s\in S$ be the north and south pole. (For definiteness, $n = (0,0,...,0,1)$ and $s = -n$.
What is $f|_N$? Well, $p|_N$ is the identity (except for at the equator), while $q$ wraps the hemisphere all the way around the sphere. But points on the equatorial $S^{n-1}\subseteq N$ get mapped to $s$.
In other words, we may think of $f|_N$ in hyper polar coordinates as $f(\vec{\theta}, \phi) = 2\phi$. To be clear, $\phi\in [0,\pi]$ measures the angle from $(0,...,1)$ to $x\in S^{n-1}$ and $\vec{\theta} = (\theta_1,....\theta_{n-1})$ with $\theta_1\in[0,2\pi]$, but every other $\phi_i\in[0,\pi],$ is the collection of angle parameters on $S^{n-1}$. (When $n = 2$, $\vec{\theta}$ is the usual $\theta$ in spherical coordinates.)
What is $f|_S$? Well, we first use the antipodal map $a$ to move all these points into the northern hemisphere, then copy $f|_S$. So, $f|_S = f|_N \circ a$. In terms of coordinates, $a(\vec{\theta}, \phi) = (\underline{\vec{\theta}}, \pi-\phi)$ where $\underline{\vec{\theta}} = (-\theta_1, \pi-\theta_2,...,\pi-\theta_{n-1})$.
(Since the anitpodal map has degree $(-1)^{n+1}$, so far this just reproduces your proof that $\deg(f_n) = 1 + (-1)^{n+1}$.)
Thus, we may describe $f$ via $f(\vec{\theta},\phi) = \begin{cases}(\vec{\theta}, 2\phi) & \phi\in[0,\pi/2] \\ (\underline{\vec{\theta}} , 2(\pi-\phi)) & \phi\in[\pi/2,\pi] \end{cases}.$
As a sanity check, this formula clearly gives a continuous $f$ away from $\phi = \pi/2$. But $\lim_{\phi\rightarrow \pi/2^-} (\vec{\theta}, 2\phi) = s = \lim_{\phi\rightarrow \pi/2^+}(\underline{\vec{\theta}}, 2(\pi-\phi))$, so this formula describes a continuous function.
Let's assume $n $ is even. In this language, your question is to find a homotopy between $f$ and a constant map. We will write down this homotopy as a composition of two homotopies. The first homotopy uses the fact that the antipodal map $\vec{\theta}\mapsto \underline{\vec{\theta}}$ is homotopic to the identity because $S^{n-1}$ is an odd dimensional sphere. Suppose $F(\vec{\theta},t)$ is such a homotopy. (For $n=2$, you can use $F(\theta,t) = \theta + t$ with $t\in[0,\pi]$)
We claim that $f_t:=\begin{cases}(\vec{\theta}, 2\phi) & \phi\in[0,\pi/2] \\ F(\underline{\vec{\theta}},t) , 2(\pi-\phi)) & \phi\in[\pi/2,\pi] \end{cases}$ is continuous.
In fact, the argument is just as it was for $f$ above: this is clearly continuous away from $\phi = \pi/2$. And at $\phi = \pi/2$, both formulas limit to $s$, so it is continuous everywhere.
At the end of this homotopy, we get a new map $f_1 = \begin{cases}(\vec{\theta}, 2\phi) & \phi\in[0,\pi/2] \\ (\vec{\theta} , 2(\pi-\phi)) & \phi\in[\pi/2,\pi] \end{cases}.$
Intuitively, this map wraps $N$ all the way around $S^n$ with decreasing lattitude mapping to even more southern lattitudes, then wraps $S$ around the sphere with decreasing lattitude moving northern.
Let's use a final homotopy to get the constant map.
The homotopy here is $G(\vec{\theta}, \phi, t) = \begin{cases}(\vec{\theta}, 2t\phi) & \phi\in[0,\pi/2] \\ (\vec{\theta} , 2t(\pi-\phi)) & \phi\in[\pi/2,\pi] \end{cases}.$ Once again, we need to check that this is continous, and again, this is obvious away from $\phi = \pi/2$.
But when $\phi = \pi/2$, both the top and bottom map send $(\vec{\theta},\phi)$ to $(\vec{\theta}, \pi t)$, so $G$ is continuous. Finally, simply note that $G(\vec{\theta}, \phi, 0) = (\vec{\theta},0) = n$, so is constant.
Best Answer
You're misinterpreting the boundary map. There is just one circle, formed by the union of the two $1$-cells. (The degree is defined from the boundary of each $2$-cell to each circle in the $1$-skeleton.) So the boundary of the $2$-cell is the sum $e_1^1+e_1^2$. Note that you have $C_2 \cong \Bbb Z$, $C_1 \cong \Bbb Z\oplus\Bbb Z$, and $C_0 \cong \Bbb Z\oplus \Bbb Z$. We have \begin{align*} \partial_2\colon C_2\to C_1\,, &\quad \partial_2(e_2) = e_1^1+e_1^2, \\ \partial_1\colon C_1\to C_0\,, &\quad \partial_1(e_1^1) = e_0^2-e_0^1, \partial_1(e_1^2)= e_0^1-e_0^2. \end{align*} You can check that this gives the homology $H_2 \cong 0$, $H_1\cong 0$, $H_0\cong\Bbb Z$, as it should.