Answer
We prove the following:
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( { - {F_1}\dfrac{{\partial g}}{{\partial x}} - {F_2}\dfrac{{\partial g}}{{\partial y}} + {F_3}} \right){\rm{d}}x{\rm{d}}y$
Work Step by Step
A graph $z = g\left( {x,y} \right)$ can be parametrized by
$G\left( {x,y} \right) = \left( {x,y,g\left( {x,y} \right)} \right)$
From here we obtain the tangent vectors:
${{\bf{T}}_x} = \left( {1,0,{g_x}} \right)$, ${\ \ \ \ \ }$ ${{\bf{T}}_y} = \left( {0,1,{g_y}} \right)$
And the normal vector:
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{{g_x}}\\
0&1&{{g_y}}
\end{array}} \right| = - {g_x}{\bf{i}} - {g_y}{\bf{j}} + {\bf{k}}$
Write ${\bf{F}}\left( {x,y} \right) = {F_1}{\bf{i}} + {F_2}{\bf{j}} + {F_3}{\bf{k}}$. By Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\bf{F}}\left( {G\left( {x,y} \right)} \right)\cdot{\bf{N}}\left( {x,y} \right){\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( {{F_1}{\bf{i}} + {F_2}{\bf{j}} + {F_3}{\bf{k}}} \right)\cdot\left( { - {g_x}{\bf{i}} - {g_y}{\bf{j}} + {\bf{k}}} \right){\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( { - {F_1}{g_x} - {F_2}{g_y} + {F_3}} \right){\rm{d}}x{\rm{d}}y$
Hence,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \left( { - {F_1}\dfrac{{\partial g}}{{\partial x}} - {F_2}\dfrac{{\partial g}}{{\partial y}} + {F_3}} \right){\rm{d}}x{\rm{d}}y$