How does adding an output resistor isolate an op amp from a capacitative load?
The resistor limits the amount of current that the op-amp has to / can sink or source, ensuring that the output transistor/s are never driven into saturation even when the op-amp is charging a completely empty capacitor (which briefly resembles a dead short to ground, which will almost certainly cause saturation).

A 'pole' is the frequency at which a filter starts have a noticeable effect; for example and very broadly, a low-pass filter with a pole at 100Hz would cause signals with a frequency of more than 100Hz to be attenuated, while frequencies of less than 100Hz would be unaffected.

Pretty much any electrical circuit exhibits some filter-like behaviour, and has poles (and their complement, 'zeroes').  A circuit made up of a resistor in series with a capacitor to ground is the archetype of a low-pass filter; it creates a pole at a frequency 1 / (2 * pi * R * C).

If an op-amp starts oscillating (usually seen as the output madly swinging between minimum and maximum as fast as it can go, at very high frequency), typically due to too much positive (self-reinforcing) feedback making it unstable, then a low-pass filter on the output will attenuate the high-frequency oscillations, which makes the op-amp more stable - less likely to start oscillating and more likely to stop oscillating if it does start.