If I use the change of variables for evaluating an integral ${\int\int}_R$ then when I use transformation why is a right-angle co-ordinate system for the new variables.
For example, to evaluate an integral the following transformation is used $u=y-x$ and $v=4y-x$ now $y-x$ and $4y-x$ are graphs not at right angles to each other which would imply $u$ and $v$ and are not perpendicular but despite this, we set $u$ and $v$ perpendicular to each other.
Yes, indeed, this coordinate transformation is not merely a rotation and scale change, but also a skewing of the angle between the axis. A square in the new coordinates was translated from a parallelogram shape in the old system.
The change of area is included in the Jacobian factor.
$$\begin{align}\iint f(x,y)\,\mathrm d\langle x,y\rangle &= \iint \begin{Vmatrix}\dfrac{\partial\langle (v-4u)/3,(v-u)/3\rangle}{\partial\langle u,v\rangle}\end{Vmatrix} f((v-4u)/3,(v-u)/3)\,\mathrm d\langle u,v\rangle\\[1ex]&= \iint \begin{Vmatrix}-4/3 & 1/3\\-1/3&1/3\end{Vmatrix} f((v-4u)/3,(v-u)/3)\,\mathrm d\langle u,v\rangle\\[1ex]&= \iint \tfrac 13 f((v-4u)/3,(v-u)/3)\,\mathrm d\langle u,v\rangle\end{align}$$
This still happens to be a linear transformation (the Jacobian factor is a scalar). Other transformations have curves in the old coordinate system become the new axis.