I asked myself the following question.
Suppose that $X_1$ and $X_2$ have densities $f$ and $g$. What is the joint density of $(X_1, X_1+X_2)$?
My effort: Based on the heuristics from Joint distribution of $(X_1,X_1 + X_2)$ that covered a discrete case.
I kind of want to do something like the following:
$ \mathbb{P}(X_1 = x_1, X_1+X_2 = x_2) = \mathbb{P}( X_1+X_2 = x_2 \mid X_1 = x_1) \mathbb{P}(X_1 = x_1) = \mathbb{P}(X_2 = x_2-x_1) \mathbb{P}(X_1 = x_1) = f(x_1) g(x_2-x_1). $
But of course all these numbers are 0 and when I try with $\mathbb{P}(X_1 \leq x_1)$ the details do not seem to work out.
The joint density of $(X_1, X_2)$ is $f_{(X_1, X_2)}(x,y)=f(x)g(y)$
It holds that $$ (X_1, X_1+X_2) = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} X_1 \\ X_2 \end{pmatrix}. $$ Notice that the matrix $\begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}$ has determinant 1.
Therefore from the transformation theorem: https://en.wikipedia.org/wiki/Probability_density_function#Vector_to_vector it follows that the joint density of $(X_1, X_1+X_2)$ is given by $$ f_{(X_1, X_1+X_2)}(x,y) = f_{(X_1, X_2)}(x, y-x) = f(x)g(y-x). $$