Not necessarily. Couldn't find that reference I mentioned, though.
$\require{AMScd}$
The CW-complex model for this phenomenon can be taken as $\pi_3(S^1 \vee S^2)$. Then this is the same as $\pi_3(\vee_{\Bbb Z} S^2)$; in the case of finitely many wedge factors, this may be identified as the space of symmetric matrices, and so this group may be identified as finitely supported symmetric matrices drawn on the grid $\Bbb Z \times \Bbb Z$. The action of $\pi_1$ is the right-shift on $\Bbb Z$, but quotienting by this action, the group is still isomorphic to $\Bbb Z^\infty$, the isomorphism given by taking the sum over each line $x + y = n$ for $n \geq 0$. The extra terms come from Whitehead products of the two factors (for instance, if $\iota_2$ is the identity map on the 2-sphere, the bracket $[\iota_2, \iota_2] \in \pi_3(S^2\vee S^2)$ is the map whose adjunction space gives $S^2 \times S^2$, and this is distinct from the spaces you would get from $\pi_3 S^2 \oplus \pi_3 S^2$.)
If all you want is a compact manifold, possibly with boundary, we are essentially finished: any CW complex of dimension $k$ embeds in $\Bbb R^N$ for large $N$ so that it has a regular neighborhood (here taken to mean codimension 0 compact manifold with boundary containing the image of the CW complex which deformation retracts onto the complex). But in particular you could just take a copy of $S^2 \times S^1$ and delete a small open ball.
The rest of this answer assumes you really care about closed manifolds.
Following your intuition to take the $S^1 \vee S^2$ idea and make it into a question about tori instead (most relevantly: a place where the universal cover splits as a comprehensible infinite wedge of the same space, in a way which is geometrically clear), here we should instead take $M = (S^1 \times S^2) \# (S^1 \times S^2)$. You may identify the universal cover as what it would look like to take the boundary of a regular neighborhood of the Cayley graph of $F_2$ in $\Bbb R^4$; precisely, at each 4-valent vertex, we add in a copy of $(S^3 \setminus 4B^3)$, the four-times punctured 3-sphere, and along each edge we add in a copy of $S^2 \times [0,1]$.
Now define a CW complex $X$, whose 1-skeleton is the Cayley graph, and 2-skeleton additionally adds two 2-spheres at each vertex. (That's not a typo: two at each.)
There is a continuous map $X \to \widetilde{M}$ given by embedding the Cayley graph as the universal cover of some appropriate wedge of two circles in $M$, with the wedge point lying on the connected sum sphere. In each $S^3 \setminus 3B^4$, at a vertex of the Cayley graph, we wedge on the two "diagonal $S^2$s", one which has the north and east necks on one side, and the other of which has the north and west necks on one side.
I claim that this map $X \to \widetilde{M}$ is a homology equivalence and hence a homotopy equivalence by Whitehead's theorem.
Here is a version of this statement that can be proved by induction. Let $X_0$ a wedge of two 2-spheres, thought of as lying above the vertex of the Cayley graph corresponding to the identity. Let $X_i$ be the subspace of $X$ obtained from $X_{i-1}$ by adding a new edge to the Cayley graph (and the corresponding 2-spheres), and $\widetilde{M}_i$ the subspace built "above this part of the Cayley graph". (Assume I have chosen my enumeration so that this exhausts $X$.)
The space $\widetilde{M}_i$ is the $2i+2$-punctured 3-sphere, and hence has $H_2(\widetilde{M}_i;\Bbb Z) \cong \Bbb Z^{2i+1}$. The map $X_i \to \widetilde{M}_i$ is injective on $H_2$, and any of the boundary spheres of $\widetilde{M}_i$ generate the cokernel of this map.
This follows from the Mayer-Vietoris sequence, adding one new 4-punctured $S^3$ at a time (containing a new pair of 2-spheres in it); the base case is clear. Under the inductive assumption and writing $P$ for the 4-punctured 3-sphere, the Mayer-Vietoris sequence gives
\begin{CD}
0 @>>> H_2(*) @>>> H_2(X_i) \oplus H_2(S^2 \vee S^2) @>>> H_2(X_{i+1}) @>>> 0\\
@. @VVV @VVV @VVV \\
0 @>>> H_2(S^2) @>>> H_2(\widetilde{M}_i) \oplus H_2(P) @>>> H_2(\widetilde{M}_{i+1}) @>>> 0.
\end{CD}
Clearly by inductive assumption the middle map is injective, and so the only map which is not clearly injective is the rightmost; its injectivity follows from one of the four-lemmas. The cokernel of $$H_2(X_i) \oplus H_2(S^2 \vee S^2) \to H_2(\widetilde{M}_i) \oplus H_2(P)$$ is isomorphic to $\Bbb Z \oplus \Bbb Z$, both terms generated by the image of $H_2(S^2)$.
In particular, because naturality and exactness identifies $\text{Im}(H_2(X_{i+1}))$ with the image of the composite $$H_2(X_i) \oplus H_2(S^2 \vee S^2) \to H_2(\widetilde{M}_i) \oplus H_2 P \to H_2(\widetilde{M}_{i+1}),$$ and the final map in the bottom row quotients by the diagonal factor of $H_2(S^2)$ in $H_2(\widetilde{M}_i) \oplus H_2(P)$, we see that we may identify the cokernel with the class given by $H_2(S^2) \to H_2(\widetilde{M}_{i+1})$. This is the right dimension-count, but not the right class; but one may use the existing 2-spheres in the image to find a homologous cycle given as a union of the boundary 2-sphere and 2-spheres in the image of $X_i$.
This proves the inductive claim.
One therefore has at the level of the colimit that $H_2(X) \to H_2(\widetilde{M})$ is an isomorphism.
Therefore, we see that $\widetilde M = \vee_{F_2 \times F_2} S^2$, with the natural translation action of $\pi_1$. Now as before we have $\pi_3(\vee_{F_2 \times F_2} S^2)$ equal to symmetric, finitely supported integer functions on $(F_2 \times F_2) \times (F_2 \times F_2)$, and the translation action acts diagonally; the quotient by the $\pi_1$-action is free on the set $(F_2 \times F_2)/(x \sim x^{-1})$ of "upper-diagonal lines".
