This is an attempt to understand Milnor's proof that it's always possible to compute a vector field's index using a non-degenerate field. The strategy is simple: take $z$ to be an isolated zero of the given field $v$, and let $U$ be a neighborhood that isolates it. Then, pick $\rho$ a bump-function that is $1$ around $z$ and $0$ outside of a slightly larger neighborhood. There's a choice of a small regular value for $v$, call it $y$, such that the new field: \begin{equation*} \bar{v}(x) = {v}(x) - \rho(x)y \end{equation*} has all it's zeros inside the smaller neighborhood, where $\rho \equiv 1$. It's then asserted that the zeros of this new field are non-degenerate. I can't, however, convince myself of that. I suppose that the need for $y$ to be a regular value for $v$ is relevant here, but I can't see how. If I compute $d\bar{v}_{z'} (h)$, for $h \in \mathbb{R}^{n}$ and $z'$ zero of $\bar{v}$, I simply get $d\bar{v}_{z'} (h) = dv_{z'}(h)$, while I would like to check that $d\bar{v}_{z'}$ is non-singular. Any enlightenment would be appreciated.
Thanks!
PS: If this information is useful, the same construction also appears as an exercise in Guillemin & Pollack's book.