Schottky transistor, Not sure I understand it?
Oli's answer is good on the mechanics of what happens : Without the diode, as increasing base current turns the transistor on harder, the transistor Vce falls below Vbe, until the transistor saturates at Vce=0.2 or even 0.05V. 

And with the diode present, as Vce falls below about 0.45V ( 0.7V minus the diode 0.25V forward voltage) the diode will start stealing the base current, preventing the transistor from saturating. (I'm not sure why Oli says this occurs at Vce=0.7V, perhaps he was using an "ideal diode" in his simulation).

But what's missing is the why:

When a transistor saturates, the base region is awash with extra carriers, and virtually no collector potential (Vce close to 0) to attract them out of the base. Therefore when you switch off the base current, the transistor remains conducting for an appreciable period of time before turning off.

Preventing it from saturation in this manner (by removing the excess base current) means it can switch off much faster, while leaving the switch-on time unaffected.

Adding this hack to 74 series logic basically tripled its speed (74S) for the same power, or allowed significantly lower power (74LS) for the same performance.