A simple answer that IMO is easy to justify using your definition of orientation goes like this.
Given any manifold $M$ and a point $p \in M$ there is a homomorphism $O : \pi_1(M,p) \to \mathbb Z_2$ and the idea is this: if $\phi : [0,1] \to M$ is a path such that $\phi(0)=\phi(1)=p$, given any basis for the tangent space to $M$ at $p$, $T_pM$ you can parallel transport that basis along the path, and you'll get a second basis for the tangent space at $\phi(1)=p$, $T_pM$. And you can ask, is the change-of-basis map from your 1st to your 2nd basis for $T_pM$ orientation-preserving -- i.e. is the determinant of that linear transformation positive? If it is, define $O(\phi)=0$, if the determinant is negative, define $O(\phi)=1$.
Fact: the path-component of $p$ in the manifold $M$ is orientable if and only if $O$ is the zero function, $O=0$. You prove it by cutting your path $\phi$ into small segments and comparing orientations within charts -- the key analytical step is the intermediate value theorem, using that determinant is a continuous function of matrices.
Of course, in this discussion "parallel transport" assumes a Riemann metric but you don't really need a Riemann metric for this argument to work. The parallel transport of vectors along a path $\phi$ simply means continuously-varying vectors such that the vector corresponding to $t \in [0,1]$ is always tangent to the manifold, i.e. elements of $T_{\phi(t)} M$. And of course if you're transporting $n$ vectors you demand that these $n$ vectors always make basis for $T_{\phi(t)}M$.
And in the case of the Moebius band, given any concrete model of the Moebius band you transport a basis along any path that goes once around the band and $O(\phi)=1$.