I've struggled a little with the following claim quoted from (first line of) this post:
The Betti numbers of $M$ is not changed after taking the orientable cover.
Is this a valid fact? Any proof?
Obviously it is false for homology groups because the $n$-sphere $\Bbb S^{n}$ is an orientable double cover of real projective space $\Bbb RP^n$. In other words the above claim says that orientable double cover removes some torsion groups of the homology group of $M$. Is this true for general smooth covering spaces $p:\widetilde{M}\to M$?
The situation is that we consider a closed $n$-dimensional non-orientable $M$ and its orientable two-sheeted covering space $\tilde M$.
You ask whether the Betti numbers of $M$ agree with Betti numbers of $\tilde M$.
The answer is no. In fact, $H_n(M) = 0$ and $H_n(\tilde M) = \mathbb Z$, i.e. $\beta_n(M) = 0$ and $\beta_n(\tilde M) = 1$.
However, the claim "Note that the Betti number is not changed after taking the orientable cover" in your linked answer only concerns the first Betti number. This is certainly true in some cases, but as Moishe Kohan comments, the Klein bottle is a counterexample.
Anyway, for the linked question it is irrelevant whether $\beta_1(M) = \beta_1(\tilde M)$. The OP only wants to prove that $\beta_1(M^3) \le 1$. In the answer it shown that this is true for an orientable $M^3$. But as Moishe Kohan comments, we have $\beta_i(M^3) \le \beta_i(\tilde M^3)$.