The Jacobian has the property
J=\frac{\partial(x,y)}{\partial(u,v)}=\frac{1}{\frac{\partial(u,v)}{ \partial (x,y)}}
so the change of variables can be written as:
\int_{S}f(x,y)dxdy=\int_{S'}f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{ \partial(u,v)}\right|dudv=\int_{S'}f(x(u,v),y(u,v))\frac{1}{\left|\frac{\partial(u,v)}{ \partial (x,y)}\right|}dudv
This is why you can first write u=u(x,y); v=v(x,y) and then find the inverse of the corresponding Jacobian. Of course, you can easily verify that the result comes out the same if you use their inverses directly, that is, write x=x(u,v);y=y(u,v) and find the corresponding Jacobian, it's just that in this particular example the first method works faster. As you can see, the dudv part remains unaffected either way.