I have 2 i.i.d Uniform random variables $\theta_{1}$ and $\theta_{2}$ on $[0, 2\pi)$. Here the problem pushes us to consider that all operations are made modulo $2\pi$.
My question is how to obtain the joint PDF of $\theta_{1} - \theta_{2}$ and $\theta_{1}$ and then obtain the conditional PDF of $\theta_{1}$ given $\theta_{1} - \theta_{2}$??.
For the second part It can be deduced from the first by writing it as the joint PDF divided by the PDF of $\theta_{1}$. But for the first part, A common idea that I saw is to derive the $\text{CDF}(x, y) = P(\theta_{1} - \theta_{2} \le x$, $\theta_{1} \le y$) and then perform derivations respect to $x$ and $y$. Is this a right approach? If yes, how to proceed with CDF especially with our uniform distribution restrictions?
You have to take into consideration the following system
$$\begin{cases} z=\theta_1-\theta_2\\ v=\theta_1 \end{cases}\rightarrow \begin{cases} \theta_1=v\\ \theta_2=v-z \end{cases}$$
The jacobian is obviously $|J|=1$ and thus
$$f_{ZV}(z,v)=\frac{1}{4\pi^2}\cdot\mathbb{1}_{(0;2\pi)}(v)\cdot\mathbb{1}_{(v-2\pi;v)}(z)$$
That is the joint distribution is defined over the following parallelogram
Knowing the joint distribution, the rest of the exercise follows as a consequence without difficulties
The fact that the uniform marginals are defined in $(0;2\pi)$ or $[0;2\pi)$ is irrelevant
EDIT: If you want to integrate in V it is very useful to define joint support in a different but equivalent manner (this is why I showed you its drawing)
$$\mathbb{1}_{(-2\pi;0)}(z)\cdot\mathbb{1}_{(0;z+2\pi)}(v)+\mathbb{1}_{(0;2\pi)}(z)\cdot\mathbb{1}_{(z;2\pi)}(v)$$