Let $B_u:=\{z\colon f(z)=u\}$ for $u>u_0:=\min f$. Let $[0,2\pi)\ni t\mapsto(x_u(t),y_u(t))$ be any smooth parametrization of $B_u$, so that
$B_u=\{(x_u(t),y_u(t))\colon t\in[0,2\pi)\}$.

(For instance, one may take $(x_u(t),y_u(t))=(\rho_u(t)\cos t,\rho_u(t)\sin t)$, where $\rho_u(t):=f_t^{-1}(u)$ and $f_t^{-1}\colon(u_0,\infty)\to(0,\infty)$ is the function inverse to the function $f_t\colon(0,\infty)\to(u_0,\infty)$ defined by the formula $f_t(r):=f(r\cos t,r\sin t)$.)

Then
\begin{equation}
F'(u)=\int_0^{2\pi}dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}, \tag{$\ast$}
\end{equation}
and hence
\begin{equation}
F''(u)=\int_0^{2\pi}dt\,\frac d{du}\Big(\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}\Big).
\end{equation}

Here, $dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}$ is the infinitesimal length element of the curve $B_u$, and $\frac{du}{|(\nabla f)(x_u(t),y_u(t))|}$ is the infinitesimal distance between the curves $B_u$ and $B_{u+du}$ near the point $(x_u(t),y_u(t))$.

We can verify formula $(\ast)$ for e.g. $f(x,y)=x^2+y^4$ and $g(x,y)=y^2$, in which case we have the following closed-form expressions: for $u>0$
\begin{equation}
F(u)=\frac{2 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{3 \Gamma
\left(\frac{9}{4}\right)}\,u^{5/4} ,\quad
F'(u)=\frac{5 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{6 \Gamma
\left(\frac{9}{4}\right)}\,u^{1/4},
\end{equation}
and the latter expression coincides with the integral in $(\ast)$, with the parametrization $(x_u(t),y_u(t))=(u^{1/2}\cos t,u^{1/4}(\sin t)^{[1/2]})$ of $B_u$, where $w^{[1/2]}:=|w|^{1/2}\,\text{sign}\, w$.