As others have answered, the transistor does not 'rush' to the on-state, conduct too much LED current, drag its supply rail down and so rush back. It travels towards a balance point, and stays there.
Risking yet another parallel, it's the same reason why pulling a spring apart in your hands sharply doesn't result in endless oscillations when you then hold your hands still and get them pulled together a bit. The spring stretches to a balance between how hard it can pull your hands together against how strongly you are holding them apart.
Back to transistors, below is a circuit that acts as what someone once called to me a 'programmable diode'. This shows an application of this 'just-on balance' idea.
simulate this circuit – Schematic created using CircuitLab
Its normal application is with a voltage across Vce supplied through a resistor somewhere up above. I'll avoid that to keep on this circuit.
Transistor Q1 turns on but the more current it conducts, the more it drags down the supply across the potential divider and so across its own base. If base voltage Vbe goes below (say) 0.7 V, Q1 isn't on at all which means the rail isn't pulled down. So Q1 finds a balance: just enough to hold its base just on enough.
This produces the function of a configurable voltage drop, with Vce kept at something like Vbemin(Ra+Rb)/Rb volts, where Vbemin is 0.6..0.7 V and particular to that transistor.