As it came out in the comments, your doubt about cellular cohomology of $\mathbb{RP}^n$ not being isomorphic to singular cohomology was because you switched the even and odd cases in $\partial$, as seen on Hatcher (current online edition) p.144.
In fact, cellular (co)homology is always isomorphic to singular (co)homology for CW complexes: Hatcher is again a good reference for this, see pages 139 for homology and 203 for cohomology. The proof for homology doesn't involve chain maps, quasi isomorphisms or chain homotopies, rather it hinges on an ad hoc diagram chasing argument, in which one uses the fact that the chain groups of the cellular chain complex are the relative singular homology gruops $H_i(X^i,X^{i-1})$. With $\mathbb{Z}$ coefficients it is true that there is a quasi isomorphism (indeed a chain homotopy) of the two complexes, since it os a general fact that for complexes of abelian groups, isomorphic homology is a sufficient condition for there to exist a chain homotopy (see here). I'm not sure whether this holds for some larger class of modules (like modules over PIDs), but for chain complexes of modules over a generic ring this is false in general, and I'm not sure if it turns out to be true in the case of singular and cellular complexes of CW complex (this is what the OP seems to be asking here).
Again, the standard proof of the equivalence of singular or cellular cohomology doesn't operate with maps at the level of chain complexes, and is similar to the proof for homology, to the extent that the bulk of the argument is just handled by using the universal coefficient theorem and the homological case, if I remember correctly. Hatcher not only shows that $H^\bullet(X;G) \simeq H^\bullet_{CW}(X;G)$, but also that the cellular cochain complex is the dual complex of the cellular chain complex, which isn't evident from the definition.
I hope all of this answers your questions. I could add details, but Hatcher is really a very good reference for all of this.
In general the canonical quotient map $q:S^n\rightarrow\mathbb{R}P^n$ is the attaching map for the top cell of $\mathbb{R}P^{n+1}$. This means that there is a cofibration sequence
$$S^n\xrightarrow{q}\mathbb{R}P^n\rightarrow \mathbb{R}P^{n+1}$$
and in particular a long exact sequence in homology
$$\dots\rightarrow H_{k+1}\mathbb{R}P^{n+1}\rightarrow H_kS^n\xrightarrow{q_*}H_k\mathbb{R}P^n \rightarrow H_k\mathbb{R}P^{n+1}\rightarrow H_{k-1}S^n\rightarrow\dots$$
If we put a CW structure on $S^n$ with two cells in each dimension $0\leq r\leq n$ then these cells are permuted by the $\mathbb{Z}_2$-action of the antipodal map. Hence the structure descends to give a CW structure of the quotient $\mathbb{R}P^n$
which has one cell in each dimension $0\leq r\leq n$.
Then in $S^n$, the boundary orientation of the two $r$-cells agrees if $r$ is even, and is opposite in the case that $r$ is odd. This means that the cellular complex $C_*S^n$ (whose homology computes $H_*S^n$) is actually exact in degrees $\neq 0,n$. On the other hand, it means that in $C_*\mathbb{R}P^m$, where the cells in each dimension are identified, the differentials are either trivial (when the orientations are opposite), or multiplication by $2$ (when the orientations agree). Hence we compute
$$\tilde H_k\mathbb{R}P^{2r}\cong\begin{cases}0&k\leq 2r\:\text{even}\\\mathbb{Z}_2&k\leq 2r\:\text{odd}\end{cases}$$
and
$$\tilde H_k\mathbb{R}P^{2r+1}\cong\begin{cases}0&k< 2r+1\:\text{even}\\\mathbb{Z}_2&k<2r+1\;\text{odd}\\
\mathbb{Z}&k=2r+1\end{cases}$$
This gives us that whenever $n=2r+1$ is odd we have an exact sequence
$$H_{2r+2}\mathbb{R}P^{2r+2}\rightarrow H_{2r+1}S^{2r+1}\xrightarrow{q_*} H_{2r+1}\mathbb{R}P^{2r+1}3\rightarrow H_{2r+1}\mathbb{R}P^{2r+2}\rightarrow H_{2r}S^{2r+1}$$
which is exactly
$$0\rightarrow \mathbb{Z}\xrightarrow{q_*}\mathbb{Z}\rightarrow \mathbb{Z}_2\rightarrow 0.$$
Hence $q_*:H_{2r+1}S^{2r+1}\rightarrow H_{2r+1}\mathbb{R}P^{2r+1}$ is multiplication by $2$. In particular this holds for the odd integer $n=3$.
Best Answer
First a reference to read, How to compute the homology of $\mathbb{R}P^n$ with cellular homology:Peter May concise course in algebraic topology page 105. he gives $\mathbb{R}P^n$ a CW structure with one $q$-cell for $0 \le q \le n$ by passage to quotients from a CW structure on $S^n$ with two $q$-cells for $0 \le q \le n$. The cells are defined by $$j^q_{\pm}(x_1, \dots, x_q) = (\pm x_1, . . .,±x_q,±(1-\sum x_i^2)^{1/2}).$$ He then computes the differentials of the cellular chains of $S^n$, to deduces that the differentials are: $$d_q[j_{+}^q]=(-1)^q d_q[j_{-}^q]=j=[j_+^{q-1}]+(-1)^q[j_{-}^{q-1}]$$ for all $q \ge 1$. We can conclude that a generator for $H^n(S^n)$ is $[j_+^{n}]+(-1)^{q+1}[j_{-}^q]$.
He then defines the cell structure of $\mathbb{R}P^n$. If $p:S^n \to \mathbb{R}P^n$ is the quotient map, he defines the cellular structure on $\mathbb{R}P^n$ by the cellular maps $j^q=p \circ j_+^q=p \circ j_-^q$. Because he choose the CW structure for $S^n$ wisely (The antipodal map acts nice on that structure), he can easily calculate the differentials for the cellular complex of $\mathbb{R}P^n$ to find: $$d[j^q] = (1 + (−1)^q)[j^{q−1}] $$
we can now deduce that the generator $[j_+^{n}]+(-1)^{n+1}[j_{-}^n]$ of $H^n(S^n)$ is sent by $p_*$ to $(1+(-1)^{q+1})j^n$. You can now conclude that $p_*$ is zero for $n$ even and is multiplication by $2$ for $n$ odd. For your other 2 questions you should dualize your chain complex and do the same kind of calculations.