OK, for what it's worth, here's how I visualize it.
As you say, a transmission line has both distributed capacitance and distributed inductance, which combine to form its characteristic impedance Z0. Let's assume we have a step voltage source whose output impedance ZS matches Z0. Prior to t=0, all voltages and currents are zero.
At the moment the step occurs, the voltage from the source divides itself equally across ZS and Z0, so the voltage at that end of the line is VS/2. The first thing that needs to happen is that the first bit of capacitance needs to be charged to that value, which requires a current to flow through the first bit of inductance. But that immediately causes the next bit of capacitance to be charged through the next bit of inductance, and so on. A voltage wave propogates down the line, with current flowing behind it, but not ahead of it.
If the far end of the line is terminated with a load of the same value as Z0, when the voltage wave gets there, the load immediately starts drawing a current that exactly matches the current that's already flowing in the line. There's no reason for anything to change, so there's no reflection in the line.
However, suppose the far end of the line is open. When the voltage wave gets there, there's no place for the current that's flowing just behind it to go, so the charge "piles up" in the last bit of capacitance until the voltage gets to the point where it can halt the current in the last bit of inductance. The voltage required to do this happens to be exactly twice the arriving voltage, which creates an inverse voltage across the last bit of inductance that matches the voltage that started the current in it in the first place. However, we now have VS at that end of the line, while most of the line is only charged to VS/2. This causes a voltage wave that propogates in the reverse direction, and as it propogates, the current that's still flowing ahead of the wave is reduced to zero behind the wave, leaving the line behind it charged to VS. (Another way of thinking about this is that the reflection creates a reverse current that exactly cancels the original forward current.) When this reflected voltage wave reaches the source, the voltage across ZS suddenly drops to zero, and therefore the current drops to zero, too. Again, everything is now in a stable state.
Now, if the far end of the line is shorted (instead of open) when the incident wave gets there, we have a different constraint: The voltage can't actually rise, and the current just flows into the short. But now we have another unstable situation: That end of the line is at 0V, but the rest of the line is still charged to Vs/2. Therefore, additional current flows into the short, and this current is equal to VS/2 divided by Z0 (which happens to be equal to the original current flowing into the line). A voltage wave (stepping from VS/2 down to 0V) propogates in the reverse direction, and the current behind this wave is double the original current ahead of it. (Again, you can think of this as a negative voltage wave that cancels the original positive wave.) When this wave reaches the source, the source terminal is driven to 0V, the full source voltage is dropped across ZS and the current through ZS equals the current now flowing in the line. All is stable again.
Does any of this help? One advantage of visualizing this in terms of the actual electronics (as opposed to analogies involving ropes, weights or hydraulics, etc., etc.), is that it allows you to more easily reason about other situations, such as lumped capacitances, inductances or mismatched resistive loads attached to the transmission line.
In a transmission line, you have an electromagnetic wave traveling along. This is a time varying electric and magnetic field. When the wave reaches a short circuit, the short circuit enforces the rule that V=0 at that location. This destroys the conditions necessary for the wave to continue traveling. Because the electric field can no longer vary with time at that location. Without this time variation, the wave cannot continue to travel.
And, as it happens, it also creates the conditions needed for the wave to reflect.
You could also consider this from a conservation of energy perspective. An electromagnetic wave has energy. It is actually a form of traveling energy. The short circuit cannot dissipate energy (when V=0, power=0). BUT, the wave cannot continue to travel, either, as previously mentioned. So, really, there is no other thing that can happen other than reflection.
You could say that when a wave in a transmission line encounters a load, any energy which is not delivered to the load MUST be reflected in order to satisfy conservation of energy. Of course, if the load is an antenna, some of the energy will be radiated into space, but that does not really change anything. The antenna is modeled as some kind of load, and the energy that is radiated into space is accounted for by a resistor in the model.
Best Answer
This video explains it so well for being so old. Bell Labs experiment of wave propagation