I want to derive the expression for the Jacobian of u and v with respect to x and y with the following considerations :
Consider a small differential rectangular element ABCD in the x-y coordinate system as shown below. This shape is mapped to quadrilateral A´ B́ C´D́ under the mapping u = u(x,y) and v = v(x,y).
Now I need to find the ratio of the area of A´ B́ C´D́ and ABCD .
Given two vectors $s = (s_1, s_2)$ and $t = (t_1, t_2)$, the area of the parallelogram with corners at $0, s, s+t, t$ is given by $$ \left|\begin{matrix} s_1&t_1\\ s_2&t_2 \end{matrix}\right| $$ and of course, this also goes for any translation of this parallelogram. So all we really need to know are the vectors $A'B'$ and $A'D'$, and insert their components into the above deteminant.
We can use the definition of derivative to get $$ B' \approx A' + dx\cdot \left(\frac{\partial u}{\partial x}, \frac{\partial v}{\partial x}\right)\\ D' \approx A' + dy\cdot \left(\frac{\partial u}{\partial y}, \frac{\partial v}{\partial y}\right) $$ and by using the fact that we are on an "infinitessimal" scale, we may put equality here.
Inserting this into the above area formula, we get $$ \operatorname{Area}A'B'C'D' = \left|\begin{matrix} dx\cdot\frac{\partial u}{\partial x} & dy\cdot\frac{\partial u}{\partial y}\\ dx\cdot\frac{\partial v}{\partial x} & dy\cdot\frac{\partial v}{\partial y} \end{matrix}\right|\\ = dx\cdot dy\cdot \left|\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix}\right|\\ = \operatorname{Area} ABCD \cdot \left|\begin{matrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{matrix}\right| $$
PS. There is a philosophical convention at play here that I personally disagree with. In my opinion, the area of a parallelogram in the plane doesn't change just because you change the coordinate grid you place on top of it (although it may seem that way if you stretch the plane in order to present the new coordinte grid in a way that looks exactly like the old one).
The only thing that changes is the actual numbers you assign to the points in the corners. So rather than "what is the area of the new parallelogram", I would ask "given the coordinates of the parallelogram in the new coordinate system, what is the actual area?"