Consider the following integral:
$$I:=\int\limits_0^1 \int\limits_0^1 (x-y)^2 \,\mathrm dx\, \mathrm dy = \frac{1}{6}\tag{1} \quad .$$
Suppose that we want to solve it via integration by substitution and thus define $w(x,y):=x$ and $z(x,y):=x-y$. The absolute value of the Jacobi-determinant equals one and so I think that $I$ can be rewritten as
$$I=\int\limits_0^1 \int\limits_w^{w-1} z^2 \,\mathrm dz\,\mathrm dw \tag{2} \quad .$$
However, we then obtain $I=-\frac{1}{6}$. Obviously the negative sign comes from the integration limits of the $z$-integration, i.e. the sign error can be fixed by swapping the limits. But is there a reason why we should swap the limits (because I think that from a simple calculations this order is correct)? Where is the mistake?
You have to decide if you are doing signed integration or not. If you are doing signed integration, you don't take the absolute value of the Jacobian, just like you wouldn't in one dimension where signed integration is customary. If you decide to use unsigned integration like in the link you provided, your substitution changes the domain from $[0,1] \times [0,1]$ to $$\{(w,z) \in \mathbb{R}^2 : w \in [0,1], z \in [w-1,w]\}.$$ Thus, by Fubini, we get $$I = \int\limits_0^1 \int\limits_{w-1}^w z^2 \,\mathrm dz\,\mathrm dw$$ and all is fine again.