I understand that V_th is obtained by opening the terminals of theoutput, so that V_out = V_in R2/(R1+R2). The problem is that I don'tknow how to calculate the current flowing into the output whengenerators are short circuited. The result should be I = V_in/R1, butthe current after R1 will split in the current flowing into R2 and thecurrent exiting the voltage divider, so this seems wrong to me. Canyou help me understand why and how to calculate the I for the opencircuit?
Well, you expressed it pretty well. That's how you do it.
$$\begin{align*}V_{_\text{TH}}=V_{\text{open circuit}}&=V_{_\text{IN}}\frac{R_2}{R_1+R_2}\tag{voltage divider}\\\\I_{\text{shorted circuit}}&=\frac{V_{_\text{IN}}}{R_1}\tag{shorted $R_2$}\end{align*}$$
If you divide the first by the second, then \$V_{_\text{IN}}\$ cancels out and the result is \$R_{_\text{TH}}=\frac{R_1\,R_2}{R_1+R_2}\$, which on the right side is the same expression for resistance that you get if you were to take the two resistors in-parallel to each other.
Since \$R_2\$ is bypassed by a short when working out the shorted circuit current, no current will occur in \$R_2\$ as the short (or wire) is infinitely less resistance. So there's nothing left over to go through \$R_2\$.
For me, what really made me understand this situation better wasn't by focusing entirely on the above approach. It's was too abstract.
What helped me more was by neither leaving the circuit open nor by shorting it, but instead by applying different resistors with finite values, in parallel to \$R_2\$. By experimenting with a few different values there, it really helped me a lot more to see better than before and to better appreciate the abstract approach (to "believe" more in it.)
So why don't you try picking values for \$R_1\$ and \$R_2\$ and then creating two or three values for a load resistance that you apply across \$R_2\$. Work out the results from that in terms of the voltage that the load experiences in these cases and compare it to what you would get using the simplified Thevenin circuit, with only one voltage source \$V_{_\text{TH}}\$, and one source resistance, \$R_{_\text{TH}}\$. It will all work out and it will help a little, I think. Try it.