Let's say there are runners on first and third with no outs. If the runner on first attempts to steal I see two outcomes:
The catcher does not throw to second. The runner advances.
The catcher throws to second, the runner on third steals home. The runner from first might not make it in time and be out, but a run scored.
This seems to be a win-win, so why don't teams try to steal in this situation very often?
Thrown baseballs move a lot faster than people. The ball will get to second and then back to home again before the runner from third can make it home, particularly as the runner on third has to delay starting until the catcher has thrown the ball to second at all.
This is not materially different than the accepted answer, just some more elaboration. In a nutshell, the options you listed are not the only possible outcomes.
The runner at third has to wait for the throw to be released by the catcher, or he/she risks the catcher not throwing to second and instead getting the lead runner in a rundown or tagged-out situation at third. So, if the runner leaves early, you get a runner on second, no runner on third, and another out. That's, objectively, worse than 1st and 3rd without the added out.
So, let's say the runner waits for the throw, that gives them less of a jump for the run to home plate. Sometimes the fielder covering 2nd base doesn't even try to tag the runner, and comes straight home with the throw, getting the runner out, with the same result as if the runner left early and got caught.
If the catcher's throw is perfect, the tag for the out at second can be applied and the fielder can be up and throwing home in almost the same instant, then you have both players caught stealing, two more outs, and no one on base.
Sometimes (less common) the catcher even throws the ball on a low trajectory and the pitcher will intercept it if he sees the player at third breaking on the throw, much like a cut-off man with a throw from the outfield.
All of these scenarios have happened during regular play.
