I've no idea how the transistor knows when to turn on in those circuits, let alone what ratio currents are between regulator and transistor.
The transistor simply turns on when the voltage on the base drops to 0.6V below the emitter. The voltage across the emitter resistor can be calculated using Ohm's law. V = IR. Rearranging the formula to make R the subject: R = V/I. The calculation for the resistor value is simply R = 0.6/I, where I is the desired current in Amps and R the resistor value in Ohms
Note that the base-emitter voltage will be much higher than 0.6V, at high currents (the transistor data sheet will give more information) so the actual current flowing through the regulator, at full load, will be higher than that required to turn the transistor on. If the transistor turns on at 0.6V but the base is 1.2V below the emitter at full load, then the regulator should have a good enough heat sink and be capable of passing double the transistor's switch on current.
The final circuit you posted uses two PNP transistors, a configuration known as the Darlington pair (look it up using a search engine for more information). Because the base-emitter junctions in a Darlington pair are connected in series, it takes double the voltage to turn it on, so the base voltage needs to be 1.2V below the emitter to turn it on.
Its the maths i struggle to get my head round. I'm trying to work out how to adjust the formula in the relationship between the transistor base resistor, and the 1 ohm ceramic power resistor. In all honesty my maths is not a strong point.
The base resistor is not essential. It can be added to ensure that the maximum base current is not exceeded when a very large current is draw, for a short period of time. It will also reduce the gain and make oscillation less likely. The circuit will work without it, but a good rule of thumb is 10 times the value of the emitter resistor. Because only a small current flows through the base of the transistor, there will only be a small voltage drop across the base resistor, so it can be ignored, for the purposes of the calculation.