Infinitesimal generator of smooth SDE

Question

Suppose we have the SDE $dX_t = \mu_t dt + \sigma_t dS_t$ where $S_t$ is a.s. continuously differentiable. More precisely, I want to put $S_t = \sum_{i = 1}^N X_ib_i(t)$ where the $X_i$ are i.i.d standard normals, $N$ is geometric, and the $b_i$'s are in $C_c^\infty(\Bbb R)$. It follows that $X_t$ is also a.s. continuously differentiable and therefore has quadratic covariation $[X_t] = 0$. By Ito's lemma, we have $$ df(X_t) = \nabla f(X_t)dX_t$$ showing that the infinitesimal generator of $X_t$ is $ \mu_t \cdot \nabla $. When $S_t$ is Brownian motion, the generator contains second order derivatives, but apparently the finite variation of $S_t$ seems to delete them for some reason. I strongly suspect that I am missing some nuance (perhaps there is some hypothesis regarding filtrations or adaptations that I am not checking), but if my claim is correct, is there some intuition explaining why we obtain a first-order operator rather than a second-order one? If I am incorrect, is there even a way to find the infinitesimal generator of $X_t$ or is the SDE ill-defined?