Impedance matching at the CMOS input
Yes, there has to be a reflection at the destination for it to work properly.

The sequence of events is this.

There is a step output, say 5v, from the driver.

The series resistor and the transmission line act as a voltage divider. Instantaneously, the transmission lines presents a 50hom impedance to the resistor, and the 5v step is divided down to 2.5v at the line. A 2.5v step propagates along the line, along with a 50mA current step, which is 2.5v/50ohms.

The step reaches the end, finds an open circuit, and is reflected. The 2.5v is reflected in phase, and adds to the incoming 2.5v, instantly raising the voltage at the end of the line to 5v. The 50mA is reflected in antiphase, and subtracts, resulting in no further current flowing at the CMOS input.

The reflected 2.5v 50mA step now travels back to the source, where it finds a 50ohm impedance, and is fully absorbed. Everything goes quiet until the next step.

If the source did not have a resistor to bring the impedance up to 50ohms, then some of that step would be reflected back down the line again, to cause trouble.

Note that this series termination method works only with a single receiver.

At the halfway point on the line, we have a dwell at 2.5v for a long time, while the step passes and returns. This is the worst possible scenario for a logic input. If we want to drive several inputs strung out on the line, like on a computer bus for instance, we must use a low impedance 5v or resistive output 10v driver (to get a full size 5v step travelling along the line), and must use a shunt 50ohm termination at the far end (to absorb it all).

Where a line has a single receiver, the series termination method is a much lower power way of achieving clean transitions.