Let $(u_1,u_2,u_3)$ be coordinates on $\mathbb{R^3}$ with the following parametrization: $$\vec{r}\left(u_{1}, u_{2}, u_{3}\right)=x_{1}\left(u_{1}, u_{2}, u_{3}\right) \hat{i}+x_{2}\left(u_{1}, u_{2}, u_{3}\right) \hat{j}+x_{3}\left(u_{1}, u_{2}, u_{3}\right) \hat{k}, \quad\left(u_{1}, u_{2}, u_{3}\right) \in D$$ Let $$\vec{r}_{u_{1}}=\frac{\partial \vec{r}}{\partial u_{1}}, \quad \vec{r}_{u_{2}}=\frac{\partial \vec{r}}{\partial u_{2}}, \quad \vec{r}_{u_{3}}=\frac{\partial \vec{r}}{\partial u_{3}}$$ s.t $B=\left\{\vec{r}_{u_{1}}, \vec{r}_{u_{2}}, \vec{r}_{u_{3}}\right\}$ is a basis for a 3D vector space $T \mathbb{R}^{3}\left(\vec{r}\left(u_{1}, u_{2}, u_{3}\right)\right)$ s.t this vector space is a copy of $\mathbb{R^3}$ so it transfer the origin of $\mathbb{R^3}$ to the point $\vec{r}(u_1,u_2,u_3)$. Let $$\tilde{B}=\left\{\vec{f}_{u_{1}}, \vec{f}_{u_{2}}, \vec{f}_{u_{3}}\right\}$$ be the dual basis for $T\mathbb{R^3}$, Find $\tilde{B}$ and show that every $\vec{A},\vec{B} \in T\mathbb{R^3}$ can be written as $$\langle\vec{A}, \vec{B}\rangle=\sum_{i=1}^{3} \sum_{j=1}^{3} g_{i j}\left\langle\vec{f}_{u_{i}}, \vec{A}\right\rangle\left\langle\vec{f}_{u_{j}}, \vec{B}\right\rangle$$ while $$g_{i j}=\left\langle\vec{r}_{u_{i}}, \vec{r}_{u_{j}}\right\rangle$$
I notice that the formula will work $\iff$ $r_{ui}\perp r_{uj}$ but why would the basis be orthogonal?
This almost-answer delves one of the fine points in the comments.
First, by a process called diagonalization an inner product like $$\langle A,B\rangle=A^{\top}GB=\sum_{i,j=1}^3 A^iB^jg_{ij},$$ where $g_{ij}=r_{u_i}\cdot r_{u_j}$, one seek the eigenvectors of $G$.
Second, allocate them in a matrix $R$ and then $R^{-1}GR$ and also $R^{\top}GR$ will be a diagonal matrices. So, by employing $R$ as a basis change then for our inner product we'll have \begin{eqnarray*} \langle A,B\rangle&=&(RR^{-1}A)^{\top}G(RR^{-1}B)\\ &=&(R^{-1}A)^{\top}R^{\top}GR(R^{-1}B)\\ &=&(R^{-1}A)^{\top}D(R^{-1}B) \end{eqnarray*} where $D$ is diagonal.
Third, this last expression will be like $$\langle A,B\rangle=\tilde A^1\tilde B^1d_{11}+\tilde A^2\tilde B^2d_{22}+\tilde A^3\tilde B^3d_{33},$$ where $\tilde A^i$ and $\tilde B^j$ are the components of $A,B$, respectively but in the basis of eigenvectors of $G$.
With this ideas maybe you were better suited to grasp for more.