I understand the concept, but what I don't
quite understand is why the reflected RF energy isn't simply absorbed
by the 50 ohm output of the transmitter after the first reflection?

In my opinion the simplest way to answer your question is that you are assuming
that the transmitter is equivalent to a 50 ohm load, which is not true because
the transmitter is instead equivalent to the series of a 50 ohm load and a
voltage generator.

A simple DC example grossly clarifies thre issue: connect a 12V battery to the
series of a 50-ohm load and another 12V battery. How much current flows through
the load? Naught (assuming the correct polarity). So no power is dissipated in
it.

73

Tony I0JX