Why care about homotopy equivalence up to $\vee S^2$ or $\# (S^2 \times S^2)$? I have read that in the study of 4-manifolds (specifically, their classification up to homotopy) that a popular/useful notion of equivalence is that up to a wedge of 2-spheres or connected sum of $S^2\times S^2$. That is, when for two smooth 4-manifolds $M,N$, we have for some $r,s$
$$M\vee_r S^2 \cong N \vee_sS^2 \,~ \text{ or } ~\, M\#_r(S^2 \times S^2) \cong N\#_s(S^2 \times S^2).$$
My questions: why is this important? What are some of the useful results supposing this notion of equivalence? Are these notions related, and do they generalize in some way?
I know of, for instance, the following fact which follows from Donaldson's theorem:

Smooth, simply-connected 4-manifolds are homotopy equivalent iff their connected sums with $S^2 \times S^2$ are homotopy equivalent.

 A: I do not know what "useful" means so I cannot comment on that.
These are amusing, primarily because the "interesting" part of 4-manifold homotopy theory is the intersection form, which here you are stabilizing in two different ways.
Say a 4-dimensional CW complex $X_\phi = \vee_r S^2 \cup_{\phi} e^4$ special, where $\phi: S^3 \to \vee_r S^2$ is some attaching map. For special CW complexes, one has the calculation $$I: \pi_3(\vee _r S^2) \cong \text{Sym}_2(\Bbb Z^r),$$ the abelian group of $r \times r$ symmetric integer matrices (isomorphic to $\Bbb Z^{r(r-1)/2}$ as an abelian group).
Using the canonical isomorphism $H^2(X;\Bbb Z) \cong \Bbb Z^r$ and $H^4(X;\Bbb Z) \cong \Bbb Z$, the map $I$ is obtained sending $[\phi]$ to the matrix $I(\phi)$ representing the cup product $H^2 \times H^2 \to H^4$.
We say an oriented homotopy equivalence of special CW complexes is one which induces the identity on $\Bbb Z = H_4$.

Theorem. Special 4-dimensional CW complexes $X_\phi$ and $X_\psi$ are oriented homotopy equivalent if and only if $\phi$ and $\psi$ are in the same orbit of the action $ \text{Sym}_2(\Bbb Z^r) \curvearrowleft GL_r(\Bbb Z),$ given by $\phi \cdot A = A^*\phi = A^T\phi A$.

Proof. If $X_\phi$ and $X_\psi$ are homotopy equivalent, choose a cellular homotopy equivalence $f: X_\phi \to X_\psi$. It quickly follows that $f$ induces an isomorphism $\Bbb Z^r = H^2(X_\psi) \to H^2(X_\phi) = \Bbb Z^r$, written $f^*$. Because $f^*$ is the identiy on $H^4$, it follows that $f^*I_\phi = I_\psi$, and hence $I_\phi$ and $I_\psi$ are in the same orbit of the $GL_r \Bbb Z$ action.
On the other hand, given an $A \in GL_r \Bbb Z$ with $A^* I_\phi = I_\psi$, observe that there is a homotopy equivalence $f: \vee_r S^2 \to \vee_r S^2$ with $f_* = A$ on $\pi_2(\vee_r S^2)$. Because $A^*I_\phi = I_\psi$, it follows that $f \phi$ is homotopic to $\psi$, and hence $f$ extends to a map $\overline f: X_\phi \to X_\psi$ which induces $f^*$ on $H^2$ and the identity on $H^4$. Since this is a map of simply connected spaces which induces an isomorphism on all homology groups and the identity on $H^4$, it is an oriented homotopy equivalence.

Any simply connected 4-dimensional manifold may be given the structure (up to homotopy equivalence) of a special CW complex. Your two operations correspond to taking the direct sum with the trivial bilinear form $(\Bbb Z, 0)$, and to taking the direct sum with the bilinear form $$\left(\Bbb Z^2, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right).$$
While it is hard to classify symmetric bilinear forms over the integers, it is plausible to classify them stably. So while you may not be able to say much about the classification of 4-manifolds up to oriented homotopy equivalence (which, topologically, correspond to nondegenerate $I_\phi$), you may be able to say something about the classification of 4-manifolds up to some kind of homotopy equivalence which is stabilized without totally destroying the intersection form.
As it turns out the wedge-with-two-spheres approach is not useful. If $(\Bbb Z^{n+r}, \psi \oplus 0) \cong (\Bbb Z^{n+r}, \phi \oplus 0)$, then in fact $(\Bbb Z^n, \phi) \cong (\Bbb Z^n, \psi)$.
