I was trying to writ metric tensor over a surface $M$ starting with the usual arc length, but I find some unclear steps.
Consider a parametric curve in $\Bbb R^n$ $$\gamma = \{ \mathbf x(t), t \in [t_0,t_1]\}$$ we approximate with small line this curve and $d\mathbf x=(dx_1,\dots,dx_n)$ is a little piece of line. We have its square length $$ds^2 = dx_1^2+\dots+dx_n^2$$ Noting $$\label{eq_1}d\mathbf x = \mathbf {\dot x} dt\tag{1}$$ (I've interpreted this as a vector equation so e.g. $dx_j = \dot x_jdt, \forall j =1,\dots,n$) we have $$l(\gamma) =\int_{t_0}^{t_1}\sqrt{\dot x_1^2 + \dots + \dot x_n^2}dt$$
Now we migrate this reasoning to a surface $M$ with local coordinate $x = (q_1,\dots,q_m)$, a parametric curve in $M$ $$\gamma = \{ \mathbf x(q_1(t),\dots,q_m(t)), t \in [t_0,t_1]\}$$ and now $$\label{eq_2}d\mathbf x = \mathbf {\dot x} dt = \sum_{j=1}^m\frac{\partial\mathbf x}{\partial q_j}\dot q_jdt\tag{2}$$
and from here everything is quite clear. My question is why in $\eqref{eq_1}$ is looking like a vector equation while in $\eqref{eq_2}$ we use the chain rule? I've tried to apply the chain rule also in $\eqref{eq_1}$ and I think the result is $$d\mathbf x = \mathbf {\dot x} dt = \sum_{j=1}^n\frac{\partial\mathbf x}{\partial x_j}\dot x_j dt$$ squaring everything $$ds^2 = \bigg(\sum_{j=1}^n\frac{\partial\mathbf x}{\partial x_j}dx_j\bigg) = \sum_{j=1}^n\sum_{k=1}^n\frac{\partial\mathbf x}{\partial x_j}\cdot \frac{\partial\mathbf x}{\partial x_k}dx_jdx_k$$ but in this case since we are in cartesian coordinates $\frac{\partial\mathbf x}{\partial x_j}\cdot \frac{\partial\mathbf x}{\partial x_k} = \delta_{jk}$ and we return to the form written in $\eqref{eq_1}$? Thank you for the help.