Transmission line theory is complicated by many factors. Here's one simplified explanation that might help you understand a little better what you're asking about.
When you send a wave down a line, it doesn't appear instantly on the other end, the electrons push each other, and in a measurable, finite amount of time the wave is present on the other end of the line.
The line has a capacitance and an inductance, though, so when you start sending the wave the cable's capacitance and inductance absorb and release that energy as the wave travels down the line.
Since all lines feature capacitance and inductance, then we have to treat them as part of the overall circuit. If we ignore them then we may lose some of our wave, or it may be altered in some undesirable way.
We could quantify the capacitance and inductance seperately, and then call each cable according to those parameters, but we've come up with a simplification called impedance.
If you send a fast wave down the line and measure the voltage and current, then apply ohms law, regardless of what's connected at the other end of the line, during the fast transition you will see 50 ohms of load on the line for the wave. This is the line's natural capacitance and inductance reacting to the AC wave you sent.
If nothing is attached to the other end, or the other end is shorted out, the wave will be reflected back.
If you attach a 50 ohm resistor on the other end, then the waveform will be completely consumed by the resistor. This is because the cable's capacitance and inductance only consumed enough current at the transmitter's voltage and waveform type to provide the wave at that same current at the output. Since you put it in at a certain voltage, and it consumed enough current for a 50 ohm load, then the voltage, and more importantly waveform shape, are preserved at the output.
If you put too small a resistor, then it'll eat the current too quickly - faster than the waveform is actually traveling on the inductor/capacitor chain that the line is made of, and this will distort the waveform and voltage, and will reflect some of the waveform back. You'll have lost energy, not everything you sent on the line will have been consumed at the other end.
If you put to large a resistor, then it won't eat the current fast enough, and again the voltage and waveform will be altered, and again some of it will be reflected back.
This leads to two problems - energy loss, and changed waveform. So if your transmitter is designed for a 50 ohm impedance, you use a 50 ohm cable, and a 50 ohm antenna, then the majority of your transmitter's energy will go into the antenna, and will be radiated into the air, with the waveform shape the transmitter intended.
If you instead replace your 50 ohm cable with a 75 ohm cable, the cable will consume less current than expected, which alters the way the waveform travels down the line, and when it's received by the 50 ohm antenna, not all of that energy will be properly coupled into the antenna.