From Vladimir A. Zorich, Mathematical Analysis II, Second Edition, pp. 187-188: We now remark that the product $JJ^{\ast}$ of the matrix $J$ and its transpose $J^{\ast}$ has elements that are none other than the matrix $G = (g_{ij})$ of pairwise inner products $g_{ij} =\langle{}\xi_i,\xi_j\rangle{}$ of these vectors, that is, the Gram matrix of the system of vectors $\xi_1,\dotsc,\xi_k$. Thus \begin{align} \mathrm{det}\,G=\mathrm{det}(JJ^{\ast})=\mathrm{det}\,J\,\mathrm{det}\,J^{\ast}=(\mathrm{det}\,J)^2 \tag{12.7} \end{align} and hence the nonnegative value of the volume $V(\xi_1,\dotsc,\xi_k)$ can be obtained as \begin{align} V(\xi_1,\dotsc,\xi_k)=\sqrt{\mathrm{det}\left(\langle{}\xi_i,\xi_j\rangle{}\right)}. \tag{12.8} \end{align}
Now let $\boldsymbol{\mathrm{r}}\colon{}D\to{}S\subset\mathbb{R}^{n}$ be a $k$-dimensional smooth surface $S$ in the Euclidean space $\mathbb{R}^{n}$ defined in parametric form $\boldsymbol{\mathrm{r}} = \boldsymbol{\mathrm{r}}(t^{1},\dotsc,t^{k})$, that is, as a smooth vector-valued function $\boldsymbol{\mathrm{r}}(t) = (x^1,...,x^n)(t)$ defined in the domain $D\subset{}\mathbb{R}^{k}$. Let $\boldsymbol{\mathrm{e}}_1,\dotsc,\boldsymbol{\mathrm{e}}_k$ be the orthonormal basis in $\mathbb{R}^{k}$ that generates the coordinate system $(t^1,\dotsc,t^k)$. After fixing a point $t_0 = (t^1_0,\dotsc,t^k_0)\in{}D$, we take the positive numbers $h^1,\dotsc,h^k$ to be so small that the parallelepiped $I$ spanned by the vectors $h^{i}\boldsymbol{\mathrm{e}}_i\in{}TD_{t_0}, i =1,\dotsc,k$, attached at the point $t_0$ is contained in $D$.
Under the mapping $D\to S$ a figure $I_S$ on the surface $S$, which we may provisionally call a curvilinear parallelepiped, corresponds to the parallelepiped $I$. Since \begin{align} & \boldsymbol{\mathrm{r}}\left( t_{0}^{1},\dotsc,t^{i-1}_{0},t^{i}_{0}+h^{i},t^{i+1}_{0},\dotsc,t^{k}_{0}\right)-\boldsymbol{\mathrm{r}}\left(t_{0}^{1},\dotsc,t^{i-1}_{0},t^{i}_{0},t^{i+1}_{0},\dotsc,t^{k}_{0}\right)\\ &\quad=\frac{\partial{}\boldsymbol{\mathrm{r}}}{\partial{}t^{i}}(t_{0})h^{i}+o(h^{i}), \end{align} a displacement in $\mathbb{R}^{n}$ from $\boldsymbol{\mathrm{r}}(t_0)$ that can be replaced, up to $o(h^i)$, by the partial differential $\frac{\boldsymbol{\mathrm{r}}}{\partial{}t^{i}}(t_0)h^{i}=:\dot{\boldsymbol{\mathrm{r}}}_{i}h^i$ as $h^i\to{}0$ corresponds to displacement from $t_0$ by $h^i\boldsymbol{\mathrm{e}}_i$. Thus, for small values of $h^i, i = 1,\dotsc,k$, the curvilinear parallelepiped $I_S$ differs only slightly from the parallelepiped spanned by the vectors $h^{1}\dot{\boldsymbol{\mathrm{r}}}_1,...,h^{i}\dot{\boldsymbol{\mathrm{r}}}_k$ tangent to the surface $S$ at $\boldsymbol{\mathrm{r}}(t0)$. Assuming on that basis that the volume $\varDelta V$ of the curvilinear parallelepiped $I_S$ must also be close to the volume of the standard parallelepiped just exhibited, we find the approximate formula \begin{align} \varDelta V \approx \sqrt{\mathrm{det}(g_{ij})(t_0)}\varDelta{}t^{1}\cdot{}\dotsc\cdot{}\varDelta{}t^{k},\tag{12.9} \end{align} where we have set $g_{ij}(t_0) = \langle{}\dot{\boldsymbol{\mathrm{r}}}_i,\dot{\boldsymbol{\mathrm{r}}}_j\rangle{}(t_0)$ and $\varDelta{}t^i = h^i, i,j = 1,\dotsc,k$.
Please explain the Eq. (12.9), I'm confusing about $\varDelta{}t^{1}\cdot{}\dotsc\cdot{}\varDelta{}t^{k}$ part.