Hams usually send just one frequency down a transmission line (as opposed to say high speed data like HDMI). So it's much simpler to analyse impedance down the line, than look at the reflections. Trying to add up the effect of multiple reflections is very complicated in the time domain.
The Smith chart is the best way to do it intuitively.
In your example, if the 300 ohm section is a half wave long, or n half waves, then at that frequency the impedance remains 50 ohms. Or if you like, the two equal and opposite reflections cancel out.
If the 300 ohm section is a quarter wave long (or 1/4 + n/2), the 50 ohms is transformed into 1800 ohms.
5 cm is about $\lambda/30$ at 200 MHz, this is 1/7 of a quarter wave, so the effect will be small. In reflection thinking, the 50-to-300 reflection and the 300-to-50 reflection are nearly at the same time, and of course have opposite sign, so they almost cancel.
In the limit, as you know from basic experience, a very short length of mismatch, like a cable joint or cheap connector, has a very small effect.
For playing with the Smith chart, I recommend this site: https://www.will-kelsey.com/smith_chart/
Here is a Smith Chart showing your exact question. Amazing.
Draw the chart for the 300 ohm line with the 50 ohm load on one end. Mark the 50 ohm point, then simply follow a constant-VSWR circle around the centre (using a compass), until you reach the desired electrical angle (pi * 5 cm / 150 cm). This is the impedance at the end of the 300 ohm line.
52.2 + j 61.9 Ohms
Or about a 3:1 VSWR. So not that insignificant.