Electronic – instant current in very long transmission line

An interconnection should be considered a transmission line if reflections are detrimental to your application. As Leon Heller says, 1/10th wavelength is a good rule-of-thumb.

However, all this depends on the nature of the signal (1/10th of what wavelength?). With sine waves, the wavelength in the cable will be about $\frac{2 X 10^8}{f}$ where f is the frequency, but will vary according to the cable type. The effect of reflections on sine waves is to cause standing waves, so the amplitude will vary along the length of the cable but will always be sinusoidal. With square waves the situation is complicated by harmonics. Although the square-wave may have a low fundamental frequency, the harmonics introduced by the edges will be significant. The effect of reflections in this case is undershoot, overshoot or ringing depending on where the mismatches are (if both $R_S$ & $R_L$ are lower than $Z_0$ for example, you will get ringing).

So do you need to match the source? No. As long as the load is perfectly matched, the wavefront will travel down the transmission line and be fully absorbed by the load. Consider a digital system with 3.3V logic for example. If you terminated both source and load you would only get 1.65V swing at the load. Terminating only at the load is termed parallel termination.

There is also a technique known as series termination in which the source is terminated but the load is left open circuit. In this case, a half-amplitude edge propagates down the line and is reflected back to the source, lifting the voltage to full amplitude as it returns until it is absorbed by the source termination. Along the line, you would see a 'stepped' signal except right at the load where you would see a clean waveform.

You are misapplying the resistor.

The problem is that the input of the receiving chip has a higher impedance than the characteristic impedance of the transmission line. Putting in a series resistor at the destination won't help much with the initial problem. Instead, you should use a shunt resistor to ground.

If you want to use a series resistor, it needs to be at the sending end. Because the driver has a low impedance relative to the transmission line, you get a discontinuity there too, and an inverted reflection results. If you put a series resistor at the driver matching the line impedance, then that is absorbed, and you eliminate the ringing caused by back-and-forth propagation of inverted reflections which is so evident on your scope.

An interesting case is a series resistor at the driver, and a high impedance receiver input without a terminator. Initially, the series resistor and line impedance form a voltage divider, and only half the applied step goes down the line. When it gets to the "open" end formed by the receiver, it reflects in phase, doubling the voltage back to the original. When the reflection gets back to the transmitter, it is absorbed by the receiver. Gradually the line voltage floats up to the drive voltage, as the situations becomes more like DC than like a step function.