This would be a good one to take in slow-motion:
- The 10k resistors want to pull the bases to the positive supply, kinda like springs, but the caps are kinda like syrup - it'll get there, but not immediately. As soon as one gets high enough to turn on its transistor, the opposite one gets jerked low through the cap and forces its transistor off.
- After the jerk low, the voltage starts to rise again, slowly, and eventually gets high enough to turn its transistor on. This jerks the first one low, and the process repeats.
The 470 resistors are required to pull up the opposite sides of the caps so that the jerk low actually does something. Using the spring analogy some more, they need to be really stiff compared to the 10k's so that we can ignore them during that part of the analysis. And that point is also convenient to tap off and feed something else.
In your case, you put an LED directly across the transistor so as to use the existing 470 pullup for double duty. It'll probably work, but it limits how far the voltage can rise at that point to the forward voltage of the LED, which limits how far the jerk can go. If the circuit still works with that little voltage swing, then you're probably good.
As for the frequency, it has to do with how long it takes the RC circuit to get from where it got jerked down to, to the base-emitter voltage of the transistors. This is not a simple RC cutoff, as in signal filtering, nor is it a trivial relation to an RC time-constant. You can either experiment with different values until you like it, or you can derive the math a couple of different ways and then solve that.