The answer is yes.
For $(u,v)\in V$, we have $g(u,v)=(x(u,v), y(u,v), f(x(u,v),y(u,v)))$ for some differential functions $x,y$. $Dg$, as a matrix, is $\left[\begin{array}{cc}\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\\\frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial u}& \frac{\partial f}{\partial x}\frac{\partial x}{\partial v}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial v} \end{array}\right]$ Let $J$ be the determinant of the submatrix consisting of the first two rows of $dg$. Direct computations show that the determinant of the other two submatrices are $J\frac{\partial f}{\partial y}$ and $-J\frac{\partial f}{\partial x}$. Since $dg$ is of rank 2, $J$ cannot be 0. This completes the proof since $\pi\circ dg$ is the submatrix consisting of the first two rows of $dg$.