Let $(V,W)$ a point in the circle of unity radius chosen in accordance with the following rules. First, let $R$ a random number uniform in $(0,1)$. Second, you choose a point $X$ on the circumference of radius $R$ just selected. Find the density of point $(V,W)$.
Well, by the inverse transformation I know that $V=Rcos\theta$, $W=Rsin\theta$, $R=\sqrt{V^2+W^2}$ and $\theta=arctan \frac{W}{V}$. So why professor writes $J=\begin{bmatrix} \frac{\partial{R}}{\partial{V}} & \frac{\partial{R}}{\partial{W}}\\ \frac{\partial{\theta}}{\partial{V}} & \frac{\partial{\theta}}{\partial{W}} \end{bmatrix}$ and not $J=\begin{bmatrix} \frac{\partial{V}}{\partial{R}} & \frac{\partial{V}}{\partial{\theta}}\\ \frac{\partial{W}}{\partial{R}} & \frac{\partial{W}}{\partial{\theta}} \end{bmatrix}$? Could you please help me to understand? In other terms, why we have the application $(R,\theta)\rightarrow(V,W)$ and not $(V,W)\rightarrow(R,\theta)$?
I don't know why one choice of $J$ seems more intuitive to you than the other choice but let me try to explain.
The process described says that the radius $R$ and the angle $X$ are chosen uniformly at random from the intervals $(0,1)$ and $[0,2\pi)$ respectively. So although it's a radius and an angle:
the point $(R,X)$ is chosen uniformly from the rectangle $(0,1)\times [0,2\pi)$.
This might be a bit counterintuitive but is completely correct from an analytic point of view. In particular we can see that the density of $(R,X)$ is $\frac{1}{2\pi}$, which means that if $A \subset (0,1)\times [0,2\pi)$, then \begin{equation}\tag{i} \mathbb{P}\bigl( (R,X) \in A\bigr) = \int_A \frac{1}{2\pi}\, drd\theta. \end{equation}
On the other hand, the point $(V,W)$ is the cartesian coordinates of a point in the disc. So:
the point (V,W) is chosen randomly - according to some density that you are asked to find - from the unit disc.
The density of $(V,W)$ is going to be a function $f(v,w)$ with the property that if $B$ is a subset of the unit disc $D$, then \begin{equation}\tag{ii} \mathbb{P}\bigl( (V,W) \in B\bigr) = \int_B f(v,w)\, dvdw. \end{equation}
Let $F : D \to (0,1)\times [0,2\pi)$ be defined by $$ F(x,y) = \Bigl(\sqrt{x^2+y^2},\arctan \frac{y}{x} \Bigr) = (r,\theta). $$ This is the function that maps the disc to the rectangle. You could start with the inverse of this function and still get the correct answer at the end; it doesn't matter as long as you understand the way the transformations work. Since $V = R\cos X$ and $W = R\sin X$, notice that $$ \mathbb{P}\bigl( (V,W) \in B\bigr) = \mathbb{P}\bigl( F^{-1}(R,X) \in B\bigr) = \mathbb{P}\bigl( (R,X) \in F(B)\bigr). $$ By (i) this is equal to $$ \int_{F(B)} \frac{1}{2\pi} drd\theta. $$ So now you want to transform this integral to put it in the form on the right-hand side of (i), i.e. transform it to $\int_B [ \dots ] dvdw$. Can you see now why the professor proceeded as they did?