Electronic – Do two stage common emitter current amplifiers add or multiply

Then let's just ignore the equations, then.

The base current doesn't cause collector current. It permits collector current.

In the PNP BJT you show in your circuit (I'm ignoring all of your circuit except for that detail), there is a hole diffusion current from the emitter into the base. This hole diffusion current can't get to the collector without crossing the base. (There is also an electron diffusion current from the base into the emitter, but let's ignore it for now.) The emitter is highly doped, so it has quite a concentration of acceptor holes. But the base is lightly doped, so it has a much smaller concentration of donor electrons. As the emitter hole diffusion current of holes attempts to cross the base to get to the collector, a small proportion of these forward injected holes combine with the lightly doped donor electrons in the base material and disappear, so to speak. This is called recombination and it causes the base in a PNP to begin to accumulate towards a more positive space charge and if that is allowed to continue, the hole diffusion current stops (it gets repelled, given time.) The solution here is for the base lead to supply more electrons to replace the donor electrons that were lost to recombination in the base material. Doing so restores the base's space charge and allows the forward injected hole diffusion current to continue as before on towards the collector. The base lead must continue to re-supply these electrons in the PNP transistor so that the space charge doesn't build up positively until it blocks that current.

In short, you MUST supply electrons into the PNP base in order to maintain the base space charge balance so that the injected hole diffusion currents can continue to cross from the emitter, across the base region, and then to the collector.

There are more details. I mentioned the electron diffusion current from the base to the emitter and there are also minor base-collector currents across that reverse biased junction, too. But the above gets most of the gist across. And does so without equations.

The left-hand transistor is being used as a diode - not only that, as a reverse-biased diode operated in breakdown. This process is notably noisy, and the 100 nF capacitor blocks the DC level from the subsequent amplification. A lot of NPN signal transistors have base-emitter breakdown voltage in the range of 6 to 9 volts, so a supply voltage of at least 12 volts is a good idea for this approach to work, although with proper transistor selection you might get away with a 9 volt battery. Note that, for instance, a 2N3904 has a minimum breakdown voltage of 6 volts, but that is a minimum value and cannot be relied on.