I understand the following proof of the Poincaré-Hopf theoerem in the case of orientable manifolds:
Proposition: Let $X$ be an oriented compact smooth manifold (without boundary). Let $v$ be a smooth vector field on $X$ with only isolated zeros. Then the sum of the indices of $v$'s zeros equals the Euler characteristic of $X$.
Proof: First we note a vector field with only isolated zeros is transverse to the zero section $Z$ of $TX$. Next, identifying the Euler characteristic of $X$ as the self-intersection number of $Z$ in $TX$, w.r.t. the orientations on $Z$ (from $X$) and $TX$, and using the fact that any vector field $v$ is homotopic to the zero section (since e.g. a vector bundle total space always deformation-retracts to the zero-section), we now have $\chi(X)$ equals the oriented intersection number of $Z$ with the image of $v$; the RHS can be identified as the sum of the indices of the zeros of $v$. QED.
My question is, can the above be generalized to the case when $X$ is non-orientable? Since after all, we can define the index of an isolated zero of a vector field with using any orientations. Also, the Wikipedia page states the theorem without the assumption of orientability. However, I'm not sure how to prove the statement without assuming $X$ is orientable. Would anyone have a suggestion as to how to approach this?