Given this number of questions all in one blob, I'll only discuss each one briefly, but this could get you started:
When connecting the rails together for a +/- voltage, will there be a reverse current through one supply? If yes, will a standard linear supply design (transformer/rectifier/LM317) handle that with no problems? Considering electrolytic caps and such.
A negative supply will need to "sink" current rather than source it. But I don't consider this a "reverse" current, its the correct way for current to flow in a negative supply.
An LM317 is a positive linear regulator. It can only source current, not sink current. For your negative supply you will need a negative regulator, like an LM337.
You always need to connect your electrolytic capacitors so that their + terminals are connected to a more positive voltage than their - terminals.
Also, if yes above: how can I home-hack a µC based ammeter that can understand current in both directions?
You will need an analog circuit to translate the sense-resistor voltage into the range of your A/D input. This is a big enough topic for a whole question of its own.
For a transformer with this pinout: pin 1-2 = 12 V, pin 3-4 = 12 V, what happens if we measure voltage between pins 1 and 3, or 2 and 4, with a multimeter? Is a 0 volt reading guaranteed, or does something else happen?
If you are still talking about two unconnected secondary coils wound on the same core primary, there is no guarantee about the dc voltage between them except by what you wire up external to the transformer. The ac voltage will be in phase for the two secondaries because it just follows the input on the primary. Obviously, by observing the "dot-location" on the two coils, you could arrange for the two outputs to be 180 degrees out of phase.
My current design has a 0.1 ohm shunt resistor (the supply is 1.2-13 volts or so, at up to 1.5 A). Is this a decent value? Lower is of course better for the output voltage, but the opamp has to amplify it more. I don't want to lose too much precision due to amplified noise!
The resistor value won't much affect the output voltage, provided you take your feedback for your regulator after the shunt resistor.
For 1.5 A through 0.1 Ohms, your resistor will be burning 225 mW. This is somewhat excessive for this function, and will cause the resistor value to drift due to self-heating. So you can either lose precision due to noise by reducing the resistor or lose accuracy due to thermal effects by keeping it large. I'd expect you could drop the resistor to 0.01 Ohms and still get good precision in your current measurement (10 bit at least) but that will depend on good analog design.
If you really want exceptional accuracy in this application you may want to look into "bulk metal foil" resistors from Vishay, which are much more stable than carbon film and other types w.r.t. thermal drift (and a bunch of other effects).
Assuming all voltages ultimately come from the same mains jack in one room, is it ever dangerous to put a multimeter set to voltage between ANY two places? For example, between the DC PSUs output and earth ground, or between earth gorund and a wall's live wire, etc?
If you are using correctly-rated probes and a correctly-rated meter, you should be able to probe mains without damaging the instrument or injuring yourself. If you are using incorrect equipment you could start a fire or electrocute yourself. Even some meters (like hardware-store models) that are labelled for 100-200-400 V are not really designed safely, so stick to reputable brands (Fluke, Keithley, Agilent, ...).
Building an AC-DC power supply inherently means working with mains, so if you don't know how to keep yourself safe while doing that, you might want to consider alternative projects until you get more experience.
First of all resistors aren't used to regulate voltages of any significant consumer.
There are several reasons for that, but the most important ones are that the resistor itself is dissipating all the dropped voltage and consuming power. That will have an impact on the battery life. The second equally important point is that resistors drop voltage but they do not provide voltage regulation! The amount of dropped voltage is dependent on the amount of current that passes through the resistor! So if you have a motor running with no load, the resistor will drop one voltage but when you put load on the motor, the resistor will drop higher voltage (assuming your power source can provide enough power and 9 V batteries aren't the best option here, especially for motors).
You can use a potentiometers and rheostats to obtain variable resistors that will give you different speeds for a motor, but the main problem with them is in general potentiometers are designed to dissipate small amounts of power and when adjusting voltage with a resistor, you'll have large power dissipation on the resistor which makes potentiometers unsuitable for directly adjusting voltages of large loads.
Also note that THERE IS ABSOLUTELY NO WAY TO USE A RESISTOR TO INCREASE VOLTAGE!!! This one is important! I'd not go too much into physics behind that here, but I think that the idea is basically equivalent of truing to produce oil by pushing your car backwards.
On the other hand, the linear voltage regulators behave like a special type of resistor which automatically adjusts its resistance (within certain range) so that the output voltage is (more or less) constant. They too dissipate the extra voltage as heat and aren't a good solution for large loads especially on battery power. Voltage output of linear voltage regulators can be controlled (on some regulators) and you can use them to control speed of a motor.
Now about the voltage drop using H-bridge: It's a bit more difficult to explain, but the main point is that when analyzing voltage coming to a motor you have basically two voltages: voltage in a single moment of time and average voltage over some time. Usually with H-bridge circuits, you're providing full instantaneous voltage to the load, but you're constantly turning the load on and off. This happens so quickly that the average voltage will look like a voltage lower than the input voltage and that way you can provide speed control for a motor by changing the time during which the motor is provided full voltage and time during which the motor has no power. The main advantage of that approach is that you are wasting very little power for voltage regulation. The transistors in an H-bridge will usually have low on resistance and when they're on, they are fully on and when they're off, they are fully off, so only little power is dissipated by them.
Another way of getting the right voltage is to use a switch-mode regulator. They are often more complicated and require more components or are more expensive if they come in same form factor as linear regulators. The good sides however make them very interesting. They can (depending on specific device) decrease or increase output voltage compared to input voltage and they waste very little energy as heat when doing so. They produce more noise on the output than linear regulators too. Anyway as far as motors are concerned and as far as I can see, there is no major benefit to use of switch-mode regulators compared to say PWM, since motors can survive short exposures to higher voltages with no problems at all (as long as the time is short enough so that the current is below the maximum rated current for the motor).
Now about that PWM motor controller: In general you'll need at least two wires to control it: ground wire to provide reference or ground voltage and a signal line. So if you're going to use an Arduino, you'll need to connect the negative sides of the controller's power supply and the Arduino together and you'll need to find the controller's signal line and drive it with PWM from Arduino.
Next, I see you mentioned stepper motors. They are usually controlled not by traditional H-bridgees but by stepper motor controllers. Basically a stepper motor has several inputs which control individual windings on the motor. You need to provide power to each winding in turn so that the motor will rotate. The speed is controlled usually not by voltage directly but by the amount of time each winding is energized. So to increase the speed of a stepper motor, you "simply" need to switch between the windings faster.
Now a little bit about the 9 V batteries: They are in general a poor choice for running any significant consumer because they are usually constricted by having 6 1.5 V cells connected in series. The cells themselves are very small and have low capacity which limits the capacity of the entire battery. This also affects the maximum current the battery can provide and since motors are significant consumers, the lifetime of a single battery will be very short. Some better options are to get say 6 AA (or C or D) cells and connect them in series for much higher capacity and higher maximum current. Another option (which could be much more expensive if you don't have the appropriate tools) would be to get a 12 V battery, such as a car battery and then recharge it or to get a 3 cell lithium-polymer battery or to get 6 cell NiMH battery.
Best Answer
A modest increase in R2 should suffice to re-bias the tube stage for 12V at the anode, P1 should be OK. Either recalculate using the load line method, or increase R2 until Va=12V, which will probably be around 200 ohms.
The output stage will still "work" as-is, but 125mA is quite a high current, so dissipating twice the power here will probably require heatsinking both FET and regulator for reliable operation.
Naturally you can't run the heater off 24V! So you still need to arrange a 12.6V or 6.3V supply for that.
One alternative would be to run both the heater and existing output stage off a regulated 12.6V supply (avoiding the increased power dissipation, though I would suggest heatsinking anyway), and a relatively low current 24V anode (B+) supply maybe using a voltage doubler then a regulator.
However this would mean you could no longer DC couple from V1 anode to Q1 grid cough sorry, gate. I would suggest AC coupling to Q1g, and providing the DC bias with a 1 Megohm "grid leak" resistor to a 6V supply (just a voltage divider with decoupling).
simulate this circuit – Schematic created using CircuitLab