Let $M$ be a $n$-manifold and let $S \subset M$ be a non-orientable $n$-dimensional submanifold possibly with boundary. Under what conditions can I conclude that $M$ is also non-orientable? Is compactness sufficient?
In particular consider the case where $M$ is a surface and $S$ is a Mobius strip.
No conditions needed.
Proof. Suppose that that $M$ is orientable, hence there is non-vanishing $n$-form $\omega$ defined on $M.$ As a result $\omega|_S$ is non-vanishing $n$-form on $S.$ Thus $S$ is orientable. Contradition. So $M$ must be non-orientable.
Other approach.
Here and here you have alternative approach to non-orientability. Namely.
We just need to use this theorem $\star$ twice.
Proof. Since $S$ is non-orientable, we get by $\star$ that there are two charts $(U_a,\phi_a),(U_b,\phi_b)$ of $S$ such as in $\star$. $S$ is open, hence $(U_a,\phi_a),(U_b,\phi_b)$ are charts of $M$ as well. Again by $\star$ we get that $M$ is non-orientable.
Remark. This equivalent condition is very convinient in proving that something is non-orientable. Generally you would have to prove that something (non-vanishing form, oriented atlas,...) doesn't exist. Here you just need to indicate two charts with some properties.