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 $f_t^{-1}$ is the function inverse to the function $f_t\colon(0,\infty)\to\mathbb R$ 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))|}
\end{equation}
and 
\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}