A rough analysis is as follows.
The A coil and the transistor Collector - Emitter capacitance in parallel with the LED capacitance forms a tuned circuit.
When the power is turned on, current starts to build up in both the A and B coils. Initially the current through the A coil is going into the LED.
Due to positive feed back from A to B (via the inductive coupling from A to B), the base current rises rapidly thus causing the collector current to also rise rapidly. When it reaches a peak, the collector current becomes steady, thus no voltage is induced into the B coil. The transistor starts to turn off thus starting a rapid change due to the positive feecback and the energy in the A coil is dumped into the LED.
Note that your claim that the back EMF is a high voltage is wrong.
If we assume that the voltage across the LED is 2 Volt and the battery is 1.5 V, the Back EMF will be 0.5 Volt. (Kirchoff's voltage Law)
Then the cycle starts again, ie. the current through A starts to rise, etc.