A compact surface is nonorientable if and only if it contains an embedded Mobius strip. I am wondering to what extent this result extends to higher dimensions - I suspect the answer is not at all.
Let $N^3$ be a closed nonorientable 3-manifold. Does there exist an embedded closed orientable surface $F \subset N$ so that the normal bundle of $F$ in $N$ is not trivial?
If $N$ is a nonorientable $3$-manifold, then there is a closed loop $\gamma\subset N$ that represents the nonorientability of $N$. This means that if you were to carry a nonzero element of $\bigwedge^3TN$ around this loop, it would change sign after its journey.
A closed regular neighborhood $\overline{\nu(\gamma)}$ of $\gamma$ is a twisted $D^2$ bundle over $S^1$. The boundary of this is an embedded Klein bottle $K$, so $N$ indeed has a closed nonorientable $2$-dimensional submanifold. However, $K$ is a two-sided surface by construction (that is, its normal bundle is trivial).
But, this is fine, and its converse is true: if $N$ is a $3$-manifold with an embedded nonorientable two-sided compact surface $S$, then $N$ is nonorientable. Take a loop $\gamma$ representing the nonorientability of $S$, which is a loop whose normal bundle in $S$ is a Mobius strip $M\subset S$. The product $M\times [-1,1]\subset N$ is not orientable, so neither is $N$. In fact, $\gamma$ gives an embedded Klein bottle like before.
Thus we have the following characterization: $N$ is a nonorientable $3$-manifold iff $N$ has an embedded two-sided Klein bottle.
Two-sidedness of the Klein bottle is important. The twisted $S^1$-bundle $S^1\mathbin{\tilde{\times}} S^1 \mathbin{\tilde{\times}} S^1$ over the Klein bottle is a closed orientable $3$-manifold, but its Klein bottle is one-sided.