Also, why does the current first reach the 2.2uF capacitor first?
Assume the initial condition is that the capacitor is uncharged, i.e., has zero volts across and both transistors are off.
In that case, the only path for current is through the resistors, capacitor, and LED. This is a high resistance path and so it will take some time for a voltage to develop across the capacitor (the base of Q1 will become more positive as the capacitor charges).
Eventually, the voltage across the capacitor will be large enough to forward bias the base-emitter junction of Q1 which will then 'turn on' and provide forward bias to Q2 which will then provide a low resistance path for current to the LED from the collector of Q2.
The voltage at the anode of the LED rises toward its nominal forward bias level which actually turns Q1 on 'harder' (positive feedback)
However, the capacitor must eventually discharge through the base of Q1 and then charge to the point that the voltage at the base of Q1 falls enough to 'turn off' Q1 and thus Q2 and the LED.
Now, the only path for current is through the resistors, capacitor and LED as before.