Q:

Does $ \int_{M} || \nabla_{M} f||^2 dS = \int_{\Omega} ||\nabla_{\mathbb{R^n}} (f \circ \Phi)||^2g d\lambda,$ hold?

Let $\Phi: \Omega \subset \mathbb{R}^n \rightarrow M$ and $M$ a euclidean manifold.
Is it then correct that 
$$ \int_{M} || \nabla_{M} f||^2 dS = \int_{\Omega} ||\nabla_{\mathbb{R^n}} (f \circ \Phi)||^2g d\lambda,$$
where $\nabla{M}$ is the gradient on the manifold and $\nabla_{\mathbb{R^n}}$ the standard gradient in $\mathbb{R}^n$? Somehow this equation is something that should hold, but I am not aware of a proof of this. In case, you know how to show it, you can restrict yourself to the case $n=2$ if you like, cause this might simplify things. 
I mean I know what the definition of the surface measure is and that we have 
$$ \int_{M} f dS = \int_{\Omega} (f \circ \Phi)g d\lambda,$$
but I am not sure how to see that the way the gradient enters the game here, does not cause any problems.
g is Gram's determinant.

A:

Assume $f$ has compact support within the image of $\Phi$. By the invariance of the integral under pullback,
$$\int_M ||\nabla_Mf||_M^2\,dS=\int_\Omega ||\nabla_Mf||_M^2\circ\Phi\,\sqrt{|g|}\,d\lambda.$$
Here $|g|$ is the determinant of the matrix with components $g_{ij}=(\Phi^*g_M)(\frac{\partial}{\partial x^i},\frac{\partial}{\partial x^j})$, and $g_M$ is the metric tensor on $M$ given by restriction of the ambient metric in $\mathbb{R}^n$.
We can also write $||\nabla_Mf||^2\circ\Phi$ as
$$||\nabla_Mf||^2\circ\Phi=(\Phi^*g_M)(\Phi^*(\nabla_Mf),\Phi^*(\nabla_Mf))=\left<\nabla_\Omega(\Phi^*f),\nabla_\Omega(\Phi^*f)\right>_\Omega=||\nabla_\Omega(\Phi^*f)||_\Omega^2.$$
Here $||\cdot||^2_\Omega$ and $\nabla_\Omega$ are the first fundamental form and the gradient operator on the Riemannian manifold $(\Omega,\Phi^*g_m)$. Thus,
$$\int_M ||\nabla_Mf||_M^2\,dS=\int_\Omega||\nabla_\Omega(\Phi^*f)||_\Omega^2\sqrt{|g|}\,d\lambda.$$

