Figuring out that "middle part" of the circuit is easier when re-drawn slightly differently. Especially the diodes and zeners are better drawn showing the voltage going down through the chain.
The zener + diode in series are used when you don't want any current to flow when the current flows in the "non-zener" direction. That's what D1 and D4 do, they prevent current from flowing.
For example when bufout is high (+5V), D3 will "zener" and drop Vz, D4 will be in forward so drop 0.7 V, Q2 can then switch on. Current could flow through D2 in the direction of D1 however, D1 blocks that so no current flows into Q1.
When bufout is low (0V) then the reverse happens, D1 and D2 conduct and D4 will block the current through D3.
bufout is either +5 V or 0 V, the voltage drop across a zener + diode + Vbe of a transistor is fairly constant, the rest is dropped across R1 or R2.
A constant voltage across R1 + NPN like that behaves as a current source. Same for R2 and Q2. That means the capacitor C1 is charged (Q1) and discharged (Q2) with a somewhat constant current. It depends on the values of the emitter resistors, supply voltage and zener voltages what that current will be.
What will be the voltage across the capacitor if it is (dis)charged with a constant current?
simulate this circuit – Schematic created using CircuitLab