Electronic – Will an oscillator be able to drive a 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.

The two signal lines are mirrored around a central reference point - often but no necessarily ground. When one is 'up" relative to the reference point the other is ":down" by an equal amount. The mean level is always zero relative to reference.

In a digital system the driver applies +v/-V for eg logical 1 and -V/+V for logical 0.

This is compared to an unbalanced system where one line may be considered the reference (often ground) line and the other swings relative to it. eg in the RS232 system logical 1 = -12V nominal and logical 0 = +12V nominal. When a 1 is sent the legs ae at -12/0 and when a 0 is sent the legs are at +12/0.

A balanced system is inherently more immune to noise. If fed via a twisted pair, noise induced into the pair is of the same amplitude in both legs and the differential signals still have the same amplitude +/- or -/+ relative to it. So noise levels can be greater in magnitude than the signal and the system will still work.

Voice telephony in standard telephone systems use analog balanced signals. Max transmit level is usually about -10 dBm where 0 dBm = 600 mV in 600 ohms = 1 mW. Customer end of line signals may be 10-20 dB below that and still work 'after a fashion". Despite this low level of signal you can have many volts of induced (usually 50 Hz or 60 Hz mains) noise on a cable pair and not hear any "hum.
Imbalance the line just slightly and ALL you will hear is hum.