On the homotopy type of $\mathbb R P^3$ and $\mathbb R P^2 \vee S^3$

As you observed, $$\pi_2(\Bbb{RP}^3) = \pi_2(\widetilde{\Bbb{RP}^3}) = \pi_2(S^3) = 0.$$ On the other hand, the wedge of two spaces $X \vee Y$ retracts onto either of the summands, by sending every point in $Y$ (or $X$, depending) to the common basepoint. In particular, $\pi_2(X) \to \pi_2(X \vee Y)$ is injective, because composing it with the map $\pi_2(X \vee Y) \to \pi_2(X)$ induced by the retraction gives the identity. So $\Bbb Z = \pi_2(\Bbb{RP}^2) \hookrightarrow\pi_2(\Bbb{RP}^2 \vee S^3)$, and that's enough to get a contradiction.