If $F_Y(y) = \mathrm{Pr} (Y\leqslant y)$ then \begin{align*}
F_Y(y) &= \mathrm{Pr}(f(X)+V \leqslant y).
\end{align*} Since $X$ and $V$ are independent, $$\mathrm{Pr}(f(X)+V \leqslant y) = \int_{-\infty}^\infty \int_{-\infty}^{y-f(x)} \phi (v) \, dv\,dF(x). $$ Then* \begin{align*}
f_y(y) &= F_Y'(y) \\
&= \int_{-\infty}^\infty \phi (y-f(x)) \, dF(x)
\end{align*} as required.

*In order to swap the integral and the derivative here I expect you will need to use dominated convergence. This might require some assumptions on $f$.

Edit: Let $$\psi(x,y) = \int_{-\infty}^{y-f(x)} \phi (v) \, dv = \int_{-\infty}^{y} \phi (v-f(x)) \, dv $$ by a change of variables. Since $v \mapsto \phi (v-f(x))$ is continuous for each $x$, if the lower bound on the above integral was finite then the Fundamental Theorem of Calculus would directly imply that $$\frac{\partial}{\partial y} \psi (x,y) =\phi(y-f(x)). \tag{$\ast$}$$ However, the lower bound is not finite, but $\phi$ decays rapidly so we might expect that $(\ast)$ is still true. Indeed, for $h>0$, \begin{align*}
\frac{\psi(x,y+h)-\psi(x,y)}{h} &= \frac1h \int_{y}^{y+h}\phi (v-f(x)) \, dv \\
&= \frac1h \int_0^{h}\phi (v+y-f(x)) \, dv.
\end{align*} Since $\phi$ is Lipschitz continuous (its derivative is bounded) it follows that \begin{align*}
\bigg \vert \frac{\psi(x,y+h)-\psi(x,y)}{h} - \phi(y-f(x))\bigg \vert &\leqslant \frac1h \int_0^{h} \big \vert \phi (v+y-f(x)) -\phi (y-f(x)) \big \vert \, dv \\
&\leqslant \frac{C}h\int_0^{h} \vert v \vert \, dv \\
&= Ch \to 0
\end{align*} as $h \to 0$. Hence, $(\ast)$ is true and this limit is uniform in $x$. Thus, \begin{align*}
F_Y'(y) &=\lim_{h\to0}\int_{-\infty}^\infty \frac1h \int_{y-f(x)}^{y+h-f(x)} \phi (v) \, dv\,dF(x)\\
&= \int_{-\infty}^\infty \lim_{h\to0}\frac1h \int_{y-f(x)}^{y+h-f(x)} \phi (v) \, dv\,dF(x) \\
&= \int_{-\infty}^\infty \psi_y(x,y) \,dF(x) \\
&= \int_{-\infty}^\infty \phi (y-f(x)) \, dF(x)
\end{align*}