You make it sound like there’s a part of a circuit that generates a signal with slow edges, and then another part that sharpens then. No, there is no circuit or part responsible for “edge sharpening”. The resistor(s) decide how much current is available for the output to switch state. The more current (smaller resistors), the faster it can switch.
Regarding the 6N137 circuit:
> the base pin is left floating.
This pin doesn’t do the same thing than on the 6N138. The 6N138 has a darlington inside, the 6N137 a comparator and AND gate. So pin 7 on the 6N137 is not a “base pin”, it just goes to the AND gate. It works as an enable/disable pin, and needs to be set high. Since it has an internal pull-up resistor, it can be left unconnected too. Check the datasheet.
> What part of the circuit here functions to sharpen the edges of the signal?
Simple: none!