I am reading Milnor's ''Topology from the differentiable viewpoint'' and there is a proof (Lemma 1 in section 6 if you have acces to the book) of this lemma:
Suppose that the vector field $v$ in $U$ corresponds to $v'=df \circ v \circ f^{-1} $ on $U'$, under a diffeomorphism $f: U \to U'$. Then the index of $v$ at an isolated zero is equal to the index of $v'$ at $f(z)$
So, he proves this for orientation preserving diffeomorphisms, and to prove it for orientation reversing diffeomorphisms it starts:
'' To consider diffeomorphisms that reverse orientation, it is sufficient to consider the special case of a reflection $\rho$.''
And then carries on completing the proof using $\rho$.
Why is it that you can just consider a reflection? It probably is very easy to see, since I find this a really good book, and he only avoids explaining trivial stuff. But I cannot seem to see it.
You can assume that $U'\subset \mathbb{R}^n$. Let $r$ be a reflection such that $ r(f(x))=f(x) $. If $f$ does not preserve the orientation $r\circ f$ does and $ind(r\circ f)'(v) _{r(f(x)) } =ind(v_x)$ since $r\circ f$ preserves the orientation. We also have $ind(r\circ f)'(v) _{r(f(x)) }=ind(r'(f'(v))_{r(f(x)) } =ind(f'(v))_{f(x)} $. Since $r$ is a reflection.