How this DIY High-Speed Probe doesn't reflect signals?

The point of this probe is that the scope has its 50Ω termination enabled. Since the scope input impedance is matched to the cable, there is no reflection from the scope. Once the load side is matched, we don't need to worry about any source-side impedance mismatch; reflections have already been suppressed.

The point of the 950Ω resistor on the input side is simply so that the scope probe presents a 1kΩ load to the circuit, which is a pretty reasonable load, versus a 50Ω load, which is decidedly unreasonable for a measurement device we're adding onto a signal.

Apparently "Art of Electronics" doesn't explain this well (I haven't read that book so I'm just taking your word for it). This type of probe is explained pretty well in the book "High Speed Digital Design: A Handbook of Black Magic", which I have read and I highly recommend. There is also this article by Dr. Howard Johnson, who is one of the authors of the book I just mentioned, which talks about this type of probe and why it's actually superior to any scope probe you can buy.

The resistor will indeed reflect the incoming signals back, but those signals exist in a few mm of IC pin + resistor pin, which have an infinitesimally small inductance. Thus those reflections will not be enough to produce any measurable overshoot.

On the other side of the resistor, a source with ~1kOhm resistance feeding a 50 Ohm cable will create a voltage divider.

On the scope side there is a potential that the cable inductance could create an overshoot because of a reflection, but there will be none as the 50 Ohm cable will match the 50 Ohm impedance of the input.