I have a question of the non-separating surfaces contained in a 3-Manifold with semi-bundle structure.

Suppose $M$ is a $3-$manifold with semi-bundle structure, this means that $M$ is a closed irreducible oriented $3-$manifold which can be doubly covered by a surface bundle over circle. In fact, $M = M_{1} \cup M_{2},$ both $M_{1}$ and $M_{2}$ are twisted $I-$bundles over a non-orientable surface, $M_{i} = \Sigma_{g} \tilde{\times} I,$ where $\Sigma_{g}$ is an orientable surface with genus $g.$ So if $\tilde{M}$ is the surface bundle which doubly covers $M,$ $\Sigma_{g}$ is the fiber surface in $\tilde{M}$. $M$ can be obtained by gluing $M_{1}$ and $M_{2}$ along their common boundary $\Sigma_{g}.$

My question is:

If $\Sigma$ is a non-separating surface embedded into $M,$ here $\Sigma$ can be either two sided or one sided (If I didn't make a silly mistake, this is equivallent to say that $\Sigma$ represents a nontrivial element in $H_{2}(M;\mathbb{Z}_{2})$ ). Further assume $\Sigma$ can be lifted to the $2-$sheeted covering surface bundle $\tilde{M},$ could we claim that the lifting $\tilde{\Sigma}$ is also non-separating in $\tilde{M}$? (In other words, is $\tilde{\Sigma}$ representing a nontrivial element in $H_{2}(\tilde{M};\mathbb{Z}_{2})$?)

Further, if $\Sigma$ is non-separating and it cannot be lifted to the double covering $\tilde{M},$ then I have the claim: there exists a non-separating surface $\tilde{\Sigma}$ in $\tilde{M}$ such that $\tilde{\Sigma}$ is the double covering of $\Sigma,$ is this claim correct? In this case, an obvious example is the fiber surface $\Sigma_{g}$ in the surface bundle covers a non-orientable surface in the semi-bundle.

I am inclined to believe that both claims are true, however, I cannot find a very clear and elegant proof. Can anyone give me some suggestions?