$\nabla(f;\psi)(u)=G^{-1}(\psi,u)J^t(f(\psi;-);u))$? Why $G^{-1}(\psi,u)$? 
Problem: Let $f:M\to\mathbb{R}$, in which M is a submanifold of $\mathbb{R}^n$ of class $C^k$ and dimension $m$.
In the natural basiss of a local parametrization $\psi:A\to\mathbb{R}^n,u\to\psi(u)$, the column matrix of coordinates of $\nabla(f)(\psi(u))$ is:
$\nabla(f;\psi)(u)=G^{-1}(\psi,u)J^t(f(\psi;-);u))=\sum_\limits{i=1}^{m}(G^{-1}(\psi,u))J(f(\psi;-);u)^i \frac{\partial}{u^i}|_{\psi(u)}$

Notation: G(\psi,u) is the Gram Matrix and $J(\psi,u)$ stands for the jacobian matrix.
Question:
I have been trying to derive the expression $\nabla(f;\psi)(u)=G^{-1}(\psi,u)J^t(f(\psi;-);u))$ but I cannot understand why $G^{-1}(\psi,u)$ is in the expression. How do I get $\nabla(f;\psi)(u)=G^{-1}(\psi,u)J^t(f(\psi;-);u))$?
Thanks in advance!
 A: Consider the local parametrization  $\psi: U\subset \mathbb{R}^m \to \Omega \subset M$ where $\Omega$ is an open set of $M$.
Since $M$ is a embedded in $\mathbb{R}^n$, the restriction of   the standard euclidean metric of $R^n$,  to $M$  gives a Riemannian metric on $M$. Let us call this metric $g:M\to T^0_2M$, concretely, let $p\in \Omega\subset M$, and $X,Y\in T_pM$ be two tangent vector of $M$ at $p$, then $g_p(X,Y) = \langle X,Y\rangle$ where $\langle\cdot,\cdot\rangle$ is the standard inner product of $\mathbb{R}^n$ (remember that you can identify $T_p\mathbb{R}^n $ with $\mathbb{R}^n$ and $T_p M$ with an $m$-dimensional linear subspace of $\mathbb{R}^n$ since $M$ is embedded in $\mathbb{R}^n$).
How do we write $g$ in the local coordinates system induced by  $u\mapsto \psi(u)$? 
$$g_{(\psi(u))} = \sum_{i,j} g_{ij}(u) \ \ d u^i|_{(\psi(u))} \otimes d u^j|_{(\psi(u))}$$ 
so we see that the coefficients $g_{i j}(u) = g_{\psi(u)}(\frac{\partial}{\partial u^i}|_{\psi(u)}, \frac{\partial}{\partial u^j}|_{\psi(u)}) =  \langle \frac{\partial}{\partial u^i}|_{\psi(u)}, \frac{\partial}{\partial u^j}|_{\psi(u)}\rangle$.
In other words the matrix $(g_{i,j}(u))_{i,j} = G(\psi,u)$ the Gramian matrix of the vectors $\frac{\partial}{\partial u^j}|_{\psi(u)}$.
Let $f:M\to \mathbb{R}$ be a regular function,  the differential of $f$ at $p$, is $df_p:T_p M \to T_{f(p)} \mathbb{R}$, we can identify it with a differential 1-form on $M$, $df:M\to T^*M$. In local coordinates, 
$$ df_{\psi(u)} = \sum_k \frac{\partial f(\psi(u)) }{\partial u^k} du^k|_{\psi(u)}.$$
The gradient of $f$ is defined as the vector field $\nabla f:M\to TM$ such that for any $p\in M$, $g_p( \nabla f (p), X)  = df_p(X)$  for any $X \in T_p M$.
This imples that*
$$ \nabla f (\psi(u)) = \sum_i \big(\sum_k g^{i k}(u) \frac{\partial f(\psi(u)) }{\partial u^k}\big) \frac{\partial}{\partial u^i}|_{\psi(u)} $$
where  $(g^{i j}(u))_{i,j}$ is the inverse of the matrix  $G=(g_{i,j}(u))_{i,j}$ (that is symmetric since $G$ it is). Now if you write the expression above for the gradient in matricial form you will obtain exactly the wanted result.
*the quicker  way to prove this is maybe to exploit that by definition of $g^{i k}$, we have that $\sum_{k} g^{i k}g_{k j} = \delta_{i j}$ the Kronecker delta, thus   by the definition of gradient for any fixed $k$ ,
$$ \langle \sum_i \nabla f^i \frac{\partial}{\partial u^i}, \frac{\partial}{\partial u^k}\rangle = df (\frac{\partial}{\partial u^k}) = \frac{\partial f}{\partial u^k}$$  Where $\nabla f^i $ denotes the $i$-unknown coefficient of $\nabla f$.  Now, the LHS is equal to $\sum_i g_{i k} \nabla f^i$ so we have that 
$$\sum_i g_{i k} \nabla f^i = \frac{\partial f}{\partial u^k}$$ 
Thus for any $j$,
$$ \sum_k\sum_i g^{j k}g_{i k} \nabla f^i = \sum_k g^{j k}\frac{\partial f}{\partial u^k}$$
$$ \sum_k\sum_i \delta_{i j} \nabla f^i = \sum_j g^{j k}\frac{\partial f}{\partial u^k}$$
$$ \nabla f^j = \sum_k g^{j k}\frac{\partial f}{\partial u^k}$$
