I need to find if it exists a topological manifold (with boundary) whose boundary is $S^1 \lor S^1.$ I think there isn't any. Indeed, any topological manifold whose boundary were $S^1 \lor S^1,$ the on the connecting point $P$ of the 8 any neighborhood minus $P$ would not be path connected. Does it work?
If not, I want to know first of all why my reasoning is not correct and only after a better solution.
Assume that $M$ is a manifold with boundary $\partial M = S^1 \vee S^1$. Then you are right, $\partial M \setminus \{P\}$ is not path connected. Why should this fact imply that $M$ is not manifold? Look at $M = \mathbb R \times [0,1)$ which has $\partial M = \mathbb R \times \{0\}$ so that if you remove a point from $\partial M$ you get a non-path-connected space. Perhaps you mean that $M \setminus \{P\}$ is not path connected which would in fact imply that $M$ (which must be two-dimensional) is not a manifold. But I do not see a proof for $M \setminus \{P\}$ not path connected in your text.
Thus Tyrone's comment gives the better strategy. If $S^1 \vee S^1$ would be the boundary of a manifold, then it would itself be a manifold without boundary. But this is not the case. The point $P$ has no neigborhood homeomorphic to an open interval.