Suppose I want to compute $$ I\equiv \int_R (x+y) \exp(x-y) \, dx\, dy$$ where the integration region $R$ is the rectangle delimited by $(a,b),\,(c,b),\,(c,d),\,(a,d)$, so namely $$ I\equiv \int_a^c dx\, \int_b^d dy\, (x+y) \exp(x-y) .$$
The obvious change of variables is $$u=x+y, \\ v=x-y,$$ but I'm struggling to see how the new integration limits on $u,v$ have to be determined.
Can you please break down the procedure?
The problem can be thought of as integrating $f(u,v)$ over a rectangle which is not aligned with the $u, v$ axes. A possible procedure is to split the integral in 3 pieces, $I_1$ for the 1st triangular region one encounters moving along a constant $v$ line that passes through a vertex, $I_2$ for the middle parallelogram and $I_3$ for the last triangle. One then has to find a way to parametrize the boundaries of such regions. Take a look at the figure below for reference ($I_1$ in green, $I_2$ in blue and $I_3$ in red), but the gist of it is to find the $u$ coordinate of each vertex of the rectangle, and then find the linear relation between $v$ and $u$. In the figure you can see a different $v$ function for each side of the rectangle.
More concretely, let the $u$ coordinates of the vertices of the rectangle be $$\{u_0=a+b,~u_1=c+b,~u_2=a+d,~u_3=c+d\}$$ (or switching $u_1$ and $u_2$ so as to have them ordered), and let us denote $v_i$ the $v$ coordinate of each point $$\{v_0=a-b,~v_1=c-b,~v_2=a-d,~v_3=c-d\}.$$ Then, after writing everything in a convenient way with the above definitions, the integrals become: $$ I_1 = \int_{u_0}^{u_1}\mathrm{d}u \int_{v_0+u_0-u}^{v_1-u_1+u} \mathrm{d} v ~f(u,v)/2$$ $$ I_2 = \int_{u_1}^{u_2}\mathrm{d}u\int_{v_2+u_2-u}^{v_1+u_1-u} \mathrm{d} v ~f(u,v)/2$$ $$ I_3 = \int_{u_2}^{u_3}\mathrm{d}u\int_{v_2-u_2+u}^{v_3+u_3-u} \mathrm{d} v ~f(u,v)/2$$ where the factor $1/2$ comes from the determinant of the Jacobian.
There might be sign errors all over the place, but the essence of the method is there.