What I think is happening is that the capacitor thats connected between ground and the output pin through that 4k7 res is slowly getting charged up each time the pulse goes low and the 555 sinks the current through the output pin. The cap is charging up from 0V to some voltage that can activate first transistor and then each time it gains 0.7V more volts its can over come two then three transistors and so on. What is confusing me is that why do they turn of completely? How does the cap empty to 0V fast?
How does the cap empty to 0V fast?
The diode across the 4k7 resistor fast discharges the capacitor that was previously slowly charged via the 4k7 resistor during the period when the 555 outputted a high pulse. The 555 output goes low after some time and rapidly discharges the capacitor via the diode.
In all other respects your simple analysis appears correct.
I second Andy's answer but would like to add (not allowed to comment yet), that the capacitor does not discharge to 0V. Instead it gets discharged down to about one diode drop (~0.7V) through the upper 1N4148 when the 555 is outputting a low signal.
To ensure that the lowest transistor does not turn on with this remaining voltage in the capacitor, there is the lower 1N4148 between ground and the emitter to lift the voltage needed to turn the transistor by also about one diode drop.