To show this extra term, here is an example problem. Let $x = u$ and $y = \sqrt{r^2 - u^2}$. Find $dx\,dy$ in terms of $u$ and $r$.
Using the Jacobian matrix J, we have that $dx\,dy = |J|\,du\,dr$, where $J = \begin{bmatrix}x_u & x_r \\ y_u & y_r \end{bmatrix}$. Plugging in values for this gives the correct answer $dx\,dy=\frac{r}{\sqrt{r^2-u^2}}\,du\,dr$.
Calculating the total differentials for $x$ and $y$ gives: $$dx = du$$ $$dy = y_r\,dr + y_u\,du=\frac{r}{\sqrt{r^2-u^2}}\,dr -\frac{u}{\sqrt{r^2-u^2}}\,du$$ Plugging in these values to $dx\,dy$: $$du\,\left[\frac{r}{\sqrt{r^2-u^2}}\,dr -\frac{u}{\sqrt{r^2-u^2}}\,du\right]$$ Where did the $-\frac{u}{\sqrt{r^2-u^2}}\,du^2$ term go using the Jacobian method? I'm guessing that we can somehow consider $du^2$ to be so small to ignore, but what if $dr$ were comparable to $du$, so that $dr\,du \approx du^2$? Why does the $du^2$ term go to zero, but not the $dr\,du$ term, and how does the Jacobian "know" to automatically ignore it?
