How would I construct the map? Once constucted, would I be right in saying that there is no Diffeomorphism to map back? As in $\mathbb{RP}^2$ a closed curve would have to have either $2$ points that represent the same $f(x)$ or one point that has $2$ values for $f(x)$? (It's not bijective)
Any advice is appreciated.
There is no global diffeomorphism from $S^2$ to $\mathbb{R}P^2$. This can be seen from the fact that their top de Rham cohomology groups are not isomorphic. Indeed, $H^2(S^2)=\mathbb{R}$ and $H^2(\mathbb{R}P^2)=0$.
As Chris Eagle pointed out, you cannot even have a continuous bijection between them, since their fundamental groups disagree. $\pi(S^2)=0$ while $\pi(\mathbb{R}P^2)=\mathbb{Z}_2$.
That said, there exists local diffeomorphisms between them. The "natural" map from $S^2$ to $\mathbb{R}P^2$ is the projection map $\pi$ with respect to the equivalence relation $x\sim -x \quad \forall x\in S^2\subset \mathbb{R}^3$, in the sense that $\pi$ is the coequalizer of the identity and the antipodal map on $S^2$. This $\pi$ is a local diffeomorphism.