Why isn't there an extra term in the jacobian to account for how much du and dv are perpendicular? I wanted to derive the formula for the multivariable change of basis in an integral on my own (for the 2 by 2 case). What I did was:
$$x=f(u,v)$$
$$y=g(u,v)$$
so $$dx = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv$$
$$dy = \frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv$$
Then, $$dx \wedge dy = (\frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv) \wedge (\frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv) = (\frac{\partial f}{\partial u}\frac{\partial g}{\partial v} - \frac{\partial g}{\partial u}\frac{\partial f}{\partial v}) du \wedge dv$$
I recognize that term as the Jacobian. Then:
$$dx\wedge dy = J(u,v) du \wedge dv$$
but I don't want to be working with bivectors, I want to work with scalars. I take the absolute value on both sides and since dx is perpendicular to dy:
$$dx dy = J(u,v) \sin(\theta) du dv $$
where $\theta$ is the angle between the two vectors. This is not the formula I learned in my undergraduate studies. How did the original formula work even if dx and dy were scalars?
 A: Long story short; you need to take the determinant in $u-v$ space, and the $-$ in the usual formula saves you.
Long story long; lets think about the extreme cases. Firstly, if $du$ and $dv$ are perpendicular in $x-y$ space, then no problem. In the other extreme case, let us consider $v(x,y)=u(x,y)+\epsilon \delta(x,y)$ where we will eventually take $\epsilon$ to be very small to investigate what happens when $du$ and $dv$ are almost colinear. We can now write
$$dx = \frac{\partial f}{\partial u}du + \frac{\partial f}{\partial v}dv=\frac{\partial f}{\partial u}du + \left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial f}{\partial \delta}\frac{\partial \delta}{\partial v}\right)dv$$
$$dy = \frac{\partial g}{\partial u}du + \frac{\partial g}{\partial v}dv= \frac{\partial g}{\partial u}du + \frac{\partial g}{\partial u}\left(\frac{\partial g}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial g}{\partial \delta}\frac{\partial \delta}{\partial v}\right)dv$$
such that
$$dx \wedge dy = \left(\frac{\partial f}{\partial u}\left(\frac{\partial g}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial g}{\partial \delta}\frac{\partial \delta}{\partial v}\right) - \frac{\partial g}{\partial u}\left(\frac{\partial f}{\partial u}\frac{\partial u}{\partial v}+\epsilon\frac{\partial f}{\partial \delta}\frac{\partial \delta}{\partial v}\right)\right) du \wedge dv$$
Then we can do some calculations, and hand wave that $\epsilon^2$ will be way too small to consider, to get
$$dx \wedge dy = \epsilon\left(\frac{\partial f}{\partial u}\frac{\partial g}{\partial \delta}-\frac{\partial g}{\partial u}\frac{\partial f}{\partial \delta}\right)\frac{\partial \delta}{\partial v}du \wedge dv+\mathcal{O}(\epsilon^2).$$
By again arguing that for very small $\epsilon$ we have $\sin(\epsilon)=\epsilon+\mathcal{O}(\epsilon^2)$ we can see that in this case actually
$$dx \wedge dy \approx J(u,\delta)\sin(\epsilon)du \wedge dv$$
So for both extreme cases we can see that your intuition for how these measures should act is actually spot on in a sense, the only thing is that the usual formula already 'protects' itself by having the difference of partial derivatives (so if the $u$ and $v$ functions are very similar, then they largely cancel), so by tacking on an additional $\sin$ you are overcorrecting.
Hope this helps, and apologies of using physicist level hand-waving of infinitesimals.
