Let me try a simpler explanation. In a transmission line like coax, there are two conductors (inner wire and 'shield')(but the shield is more than that, it actually carries an equal but opposite current than the inner wire). There is some capacitance between the two conductors (like a capacitor) and also some inductance (like an inductor or coil). These conspire to limit current flow for ac signals. Suppose you have an infinitely long piece of ideal RG58. If I suddenly hook up a 100 volt battery to the two conductors at one end, I will find that 2 amps flows out of the battery. Noting that volts divided by amps is the definition of 'ohms', we say say that RG58 has an impedance of 50 (100 volts divided by 2 amps) ohms. But note that nothing gets hot, no energy is lost, unlike in a resistor. Now for a finite length line, it's more complicated. Suppose the far end is open circuit. When you hook up the battery, a 100 volt/2 amp pulse races down the line at the speed of light, reflects at the open end, races again at the speed of light back to the battery, which now stops putting out the 2 amps. You're left with no current, and 100 volts between the conductors - the usual DC result. And since it all happens at the speed of light you need fairly sophisticated equipment to see it, although a good oscilliscope is sufficient. One last thing- if I put a 50 ohm resistor across the conductors at the far end, all of the energy is turned into heat, and there are no reflected pulses at all, and the battery will continue to put out 2 amps. We say the load is matched to the cable. Now an antenna is not a resistor, but as far as the coax is concerned it could be. A matched 50 ohm antenna will not get hot, but rather will radiate energy. The energy radiated is the current squared times 50 ohms - just like the power converted to heat in a resistor. So the coax cannot tell the difference.
Probably more than you cared to know.