 As Q2 base-emitter junction is reverse-biased, it does not conduct, so all the current from R2 goes into C1.Simultaneously, C2 that is fully discharged and even slightly charged to 0.6 V (in the previous State 2) quickly charges via the low-value collector resistor R4 and Q1 forward-biased base-emitter junction (because R4 is less than R2, C2 charges faster than C1).Thus C2 restores its charge and prepares for the next State C2 when it will act as a time-setting capacitor.