When the control voltage goes high why does the base of Q2/Collector of Q1 rise to 0.8V, and why does the base of q1 also rise by that same value?
Because the is a pn junction between the base of Q1 and the input (essentially a diode), the voltage at the base of Q1 will never be more than somewhere around 0.6-0.7V above the input. For a similar reason, the collector of Q1 and the base of Q2 will never be more than 0.6-0.7V below the base of Q1.
This means that when the emitter of Q1 is low, the base of Q2 will also be low.
When the emitter of Q1 is high, the base of Q1 will rise to about 1.2-1.4V. There will be one diode drop between the base and collector of Q1, and one diode drop between the base and emitter of Q2.
If the circuit seems strange at first, it is probably because when the input is high, the base-emitter junction of Q1 is reverse biased. That is not normal for CE, CB or CC amplifiers. But this transistor in not being used in as a normal CE, CB or CC amplifier. It's function is basically that of two diodes. However it has an advantage over two discrete diodes. In discrete diodes, when current switches from one diode to the other, there is a longer reverse recovery time. Minority carriers in the diode need to be cleared. With the transistor used in place of two diodes, carriers in the base do not need to be cleared. They simply change direction regarding whether they are flowing to/from the emitter to flowing to/from the collector. Hence using a transistor this way allows faster switching than using two discrete diodes. (Hence TTL or transistors transistor logic, replaced DTL, or diodes transistor logic.)