The devil's in the details, even if they are printed.
The trick is in the fact that the transmission-line is of characteristic 75 Ohm, and the excitation side is also 75 Ohm. Were that end very low source impedance it would tend much more to wanting a node (and as such forcing the need for an adjusted frequency = half the allowable frequencies), were it a very high impedance it would tend more to wanting an anti-node and again half the allowable frequencies.
The fact it is the exact characteristic makes that end reflection-less and in effect not a defining factor in the orientation of the standing wave. As such the open end gets to "dictate" the orientation on its own and the number of allowable frequencies for standing waves on the same line gets doubled.
First start by reading here and and in particular the section "Single-source transmission line driving a load"
Why is max power is transferred when the characteristic impedance of a transmission line is equivalent to the impedance of a load.
Well that is not exactly true. You should say "equivalent to the Real part of the impedance of the load."
You should know that there are 3 passive elements: Resistors, Capacitors and Inductors.
Of those only the Resistor can dissipate power because it has a Real value impedance.
$$ Z_R = R $$
Capacitors and Inductors are reactive components and cannot dissipate power (we're talking about ideal components here). Their impedance only has an imaginary part
$$ Z_C = 1/jwC $$
$$ Z_L = jwL $$
That j makes these imaginary.
These reactive componets can only influence the amplitude and phase relation of a signal. Since they cannot dissipate power no power is lost in these components.
An (ideal) transmission line can be seen as a distributed network of Capacitors and Inductors, so no resistors ! The characteristic impedance of a transmission line tells us something about the relations between amplitude, phase, currents and voltages of the waves traveling through it.
In the middle of a transmission line the wave traveling through it "sees" the same characteristic impedance in front and behind. It cannot dissipate into these impedances as they are reactive, they cannot dissipate power.
However at the end of the transmission line at the load, the characteristic impedance ends and turns into a real impedance. The amplitude and phase relations are not changed when the load impedance has the same value as the characteristic impedance of the transmission line. So the wave travels into the load as if nothing has changed. If there was a difference, then part of the wave would reflect.
In the load the wave cannot travel further but since the impedance is real it is dissipated and turned into heat.
Best Answer
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.