You are missing the fact that a transmission line is not a resistor. A 50 Ω line, terminated in a 50 Ω resistor, 'looks like' a 50 Ω load to the source driving it.
A 50 Ω line or load both define a ratio of voltage to current. However, for the line, it defines the ratio of the voltage wave to current wave that propagates along the line. The line itself has (ideally) no resistance.
There are several ways to make a source have an output impedance of 50 Ω, one way is to have a voltage source followed by a 50 Ω series resistor.
Let's say the 50 Ω source drives a 50 V step into a 50 Ω resistive load. That will need a 100 V step from the voltage source, because of voltage division between its series resistor and the load. The load current suddenly rises to 1 A, and the load voltage to 50 V.
Instead, let's put a length of 50 Ω transmission line between these. When the step occurs, the ratio of voltage wave in the line to the current wave will be 50 Ω. A voltage of 50 V will appear across the line, and a current of 1 A start flowing into it from the source. A 50 V voltage wave and a 1 A current wave set off down the line. When they reach the far end, they find a 50 Ω load. The load voltage now rises to 50 V and its current to 1 A. There has been no problem in meeting both the voltage and current from the line, they are in the correct ratio, so all the boundary conditions are met and no reflection is generated.
If instead the load had been something different, say 100 Ω, or an open circuit, or a short circuit, the current and voltage in the load would not have matched that arriving along the transmission line, and a reflected wave would have been generated to make up the differences.