I have a function on the form $$g(y) = \int_{-\infty}^{\infty}{e^{-v^2}f(y-v)dv}$$ I know that $g(y)$ is linear around $0$, $g(y\approx 0)\approx yG$, and I am interested in finding this gradient $G$.
For this reason I thought that the integrand could be expanded around $y=0$. In general, am I allowed to do that for a convolution? Or should I expand around $y-v=0$?
Yes, you should expand around $y=0$ if you're interested near there! With suitable reasonable restrictions the expansion, which is really just a derivative with respect to $y$ at 0, commutes with the integral (see http://en.m.wikipedia.org/wiki/Differentiation_under_the_integral_sign for example) so everything works out.
$$G= \int e^\cdots \partial_y f(y-v) \text{ at }y=0$$
Incidentally a trick which is sometimes useful is using $\partial_x f(x-y)=-\partial_y f(x-y)$ so that you can integrate by parts, or similar.