What it means by 6mA at 5v, is that at 6mA, the output voltage level of that output will be fairly close to 5v (or VCC). The more current you source out of an output, the lower that voltage will be. This is called Voltage Droop, or Sag. Same goes for sinking current, with the voltage level rising, so instead of it being 0.0 or 0.1, at 20mA it might be 1v.
Notice that at a VCC of 4.5V, Voh (Voltage Output High), at Ioh (Current output high) -20µA (or -0.02 milliamps, the - indicating sourcing current) the typical is 4.499 volts. While at -6 mA, the typical is 4.3? Your output just lost 0.2v with only 6 mA of current draw. Anything more, and that drop will increase an untested amount. Some datasheets actually have a graph for this. This one does not. So you can draw more than 6mA, up to the max Io output current of 35mA (plus or minus) on a single output, but the voltage level will change to the point where it might not be useful to you. Ex. At 35mA, you might not be able to light an led.