What causes the reflection? Why doesn't it occur with other
frequencies? (if it doesn't)
Reflections occur at all frequencies when there is a mismatch in impedances. At low frequencies, such as audio, these reflections are difficult to see but they are there all the same. Reflections are generally said to be significant when the frequency is high enough AND the interconnection between sender and receiver is long enough. Somewhere in the order of about a tenth of a wavelength or bigger is a general rule of thumb.
At 20 kHz, the wavelength (in 100% speed of light cable) is about 15 kilometres and if you had a cable of about 1.5 km length you might start to see the effect of reflections.
However, if you had a 100 MHz transmitter, you might see the effect of reflections at 300 mm.
Consider a battery and a lightbulb. The lightbulb is connected to the battery with a switch. The battery and switch are at one end of a lossless 10 km cable and the bulb is at the other end. When the switch closes, how much current is drawn from the battery? - how can the battery know how much current to supply in that instant? The answer is it can't - it supplies what the cable demands and, for a 50 ohm cable an appropriate current is supplied. If the voltage were 10V then the current would be 200mA.
This travels down the cable (at a power of 2W) until it hits the lightbulb. The bulb may have an impedance of (say) 100ohms - it only wants 100mA at 10V but it gets 200mA - there is a mismatch and the excess power gets reflected back up the cable to the battery and switch. This power can't be dissipated in the battery so it gets relfected back and forth. Of course, cable has real losses and these eat away at this reflection and the system stabilizes with 100mA flowing down the cable. This is a simplifed explanation.
Does this help you understand?
What happens when a very stiff source drives a transmission line? You get reflections, and those reflections occur from both ends. If the source is not terminated any mild imperfections in the cable or termination at the RX end will cause reflection that then reflect with 100% efficiency back down the cable and cause problems.
The issue then arises, how do I terminate the Tx and the Rx without killing the amplitude by 1/2?, because after all you are forming a voltage divider.
The easy approach is to put a termination (series resistor) inside the amplifier right after the ideal very stiff driver and before it hits the cable. To compensate for the loss you feed-back the voltage signal from the driven end of the cable and cause the driver to drive harder to compensate for the voltage divider termination.
More sophisticated drivers actually design the differential resistance of the amplifier to be matched accordingly and are thus more efficient. Of course you can't drive to DC with that approach but often that is not the issue.
Of course this means that any given driver must then be designed for an explicit termination impedance.
Upon edit: I forgot to mention the simplest case, where you design it to drive at 2X the voltage and have a builtin terminator (series) resistor. This is less power efficient and has the side effect that an non loaded amplifier put out 2X the voltage.
Best Answer
Loading refers to how your measurement circuit affects the attached device. If e.g. you want to read a sensor that has a big internal resistance Ro any current drawn will create you additional voltage error equal to io*Ro. Drawing as little current as possible lowers the sensor loading thus providing more accurate readout.