Manifold without point homotopy equivalent to wedge of $2$-spheres.

The answer is: if and only if $\pi_1(F)=1$.

By cellular approximation theorem we have $\pi_1(F)=\pi_1(F')$, so if $F'$ is a bouquet of spheres, our assumption is satisfied.

In the opposite direction, by $\pi_1(F)=1$ we have $H_1(F)=H^3(F)=0$ using Poincare duality. Therefore, by universal coefficient theorem, $H_2(F)$ has no torsion and $H^1(F)=0$. So, using Poincare duality again, we have $H_3(F)=0$.

By extision, $F$ and $F'$ have the same homology except $H_4$, so $\pi_2(F')=\mathbb Z^k$ for some $k$, and we may to construct a map $\vee_k S^2 \to F'$ that will be an isomorphism on $\pi_2$. Such a map will be also an equivalence on homology, so we have a desired statement.