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.
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.
Best Answer
You need to know what the maps are. Let $C_0,C_1$ and $C_2$ be the free groups on the cells. Then you have
$$0\rightarrow C_2\rightarrow C_1 \rightarrow C_1\rightarrow 0$$ $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0$$
and $$\partial_2 (n)=(2n,0)$$ $$\partial_1(n,m)=0$$ so for the homology groups we get $$H_2,H_1,H_0$$ $$0,\mathbb{Z}_2\oplus\mathbb{Z}, \mathbb{Z}$$
Now to get the cohomology we have to find the maps for
$$0\rightarrow \operatorname{Hom}(C_0, \mathbb{Z})\rightarrow \operatorname{Hom}(C_1, \mathbb{Z})\rightarrow \operatorname{Hom}(C_2, \mathbb{Z})\rightarrow 0$$ $$0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\oplus \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0$$
So $$\delta_0 f (a,b)=f(\partial_1(a,b))=0$$ thus $\delta_0=0$. $$\delta_1 g (n)=g(\partial_2(n))=g(2n,0)$$ By choosing an appropriate basis for $C^1$ we can think of this map as $(a,b)\mapsto 2a$.
Thus for cohomology we get
$$H^0,H^1,H^2$$ $$\mathbb{Z}, \mathbb{Z}, \mathbb{Z}_2$$
I hope this makes sense and that it helps !