Reducing transistor switching time with a resistor from base to a negative voltage
In the image you posted, when the switch is on, the base emitter junction is forward biased and charge carriers flood the region. When sufficient charge carriers exist, the transistor will go into saturation. When the switch is turned off, in order for the transistor to go into cutoff, the bountiful quantity of carriers in the base emitter junction need to leave. The longer it takes for them to leave, the longer it takes for the transistor to get out of saturation mode and into cutoff. 

Without the resistor R there, the carriers basically only have one way out when the switch is open, through the base emitter junction. This will take time.

With R, the charges can get out through R and go to ground. Again, this will take time but will be faster than letting the base emitter get rid of the carriers by itself.

The question you linked too, makes the switching time even faster, by adding a speedup cap across the base resistor, and by using a baker clamp ( diode across base to collector). The baker clamp will limit the number of carriers that enter the region, and the speedup cap will provide a quick negative pulse (reverse bias the base emitter) to help speed the carriers out.