Think about the signal on the transmission line as being made up of a forward-travelling wave and a reverse-travelling wave.
At any given point on the line, the ratio of voltage to current in each of these two independent waves is determined by the characteristic impedance \$Z_0\$ of the line.
The forward travelling wave was initially created by the power source driving the line.
The reverse-travelling wave only exists because something caused reflections of the forward wave to return back up the line. Reflections only occur when there's some change in the line geometry, a circuit attached to the line, or a mismatched termination.
If voltage source truly sees different impedances down the line,
Your formula for \$Z_{in}\$ applies to one specific situation: There is a mismatched termination on an otherwise ideal line and you want to know the ratio of voltage and current (due to the combination of forward and reverse waves) at some location up the line from the termination. It accounts for the fact that the phase of the forward wave is different at this point on the line then where it encounters the load, and that the phase of the reverse wave has also changed as it traveled back down the line to where you're measuring.
It doesn't mean that the characteristic impedance of the line is changed in any way: The voltage/current ratio in the forward wave is still determined by \$Z_0\$ and the voltage/current ratio in the reverse wave is still determined by \$Z_0\$. And if there isn't a discontinuity or geometry change at this point in the line, there won't be any additional reflection created because \$Z_{in} \ne Z_0\$ at this point on the line.
... then why would it deliver any power to the load at all since eventually somewhere on the line the impedance would be infinity?
The apparent input impedance will only go to infinity if the load has reflected 100% of the energy in the signal (\$|\Gamma_L|=1\$). In this case, in fact there can't be any energy delivered to the load, because all of the energy is reflected back in the reverse wave.
When you match to complex valued loads the matching for zero power reflection states that the impedance seen from your complex-valued load (\$100+50j\$) has to be its complex conjugate (\$100-50j\$).
This is because that way we would be satisfying the max. power theorem, and, at the same time, getting rid of the imaginary part of your load.
Best Answer
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.