Voltage – Stepping Down Voltage from 120 VAC to 110 VAC or 117 VAC

1: Yes, you can do that. Essentially, that's how power supplies work. They can handle multiple parallel networks within their current capacity. As for the heatsink, that depends on the regulator, the current draw, the ambient temperature, how efficient it is, etc. It's not a simple yes or no.

2: The capacitors depend on the regulator as well. Some require them all the time, some only require them depending on the input or output conditions, some never require them. The NTE1960 you linked to does not have an extensive datasheet, but is pretty similar to the LM7805. The capacitors are pretty much required for stable use. But these are linear regulators. Not efficient and they convert wasted energy into heat. Going from 12v to 5v, at say 700mA which is the high end for the RPI, that means 12 - 5 = 7v * 700mA = 4.9 Watts of energy being converted into heat. A heatsink would be required.

A Switching regulator is more efficient, in terms of both energy and heat. The OKI-78SR component you chose is a Switching Regulator. It shows that it would not need a heatsink in that same situation (Not in the engine compartment though, that's a different story). It is also a complete module, including the capacitors and the resistors it needs. It would be better.

3: A Car USB regulator would work just fine for your case, as long as the draw on it is under it's maximum. Some are 500mA, some are 1A, or better or in between, but some can't actually supply the amount of current it says it should, so you would need to test. The Model B has a 700mA draw/limit, the Model A is 500mA. Most of these usb regulators are switching supplies, and for your purposes, a car usb adaptor would be exactly like the OKI-78SR. At 4 bucks for the OKI-78SR (plus shipping) compared to a few bucks for a car USB adaptor, it really just depends on which you can get easier. Even retail, you can get a decent car one at any convenience or auto store for 10 bucks.

You could even gut the car USB adaptor for the board inside. Those things are so small now they are smaller than a car cigarette lighter, with the case, and the size of an SD card without the case.

To me, Kirchoff's Laws are scientific wording of what should be common-sense observations of electric circuits. Unfortunately, common sense isn't as common as we might like.

KVL says that the total of the voltage drops in a circuit must equal the supplied voltage. If that was not true, we would have some voltage across a wire which would result in approximately infinite current in that wire - but KCL insists that the current is the same at all points in a simple series circuit, so we can't have a huge current at one point in the circuit.

If you connect a device that normally requires 4 volts across a 6 volt battery, sufficient current will flow to make the resulting circuit comply with KVL. The voltage across the "4 volt device" will rise, and the output voltage of the battery will fall (due to a voltage drop across the internal resistance of the battery) such that the voltage across the battery and the device are equal. This may result in the destruction of the 4 volt device, if it cannot withstand the extra voltage and current.

Regarding your second point: the voltage drop across a resistor will depend on its resistance, and on the current passing through it, in accordance with Ohm's Law.

For your example of a 10 Ohm resistor across a 30 volt supply, the current through the resistor will be 3 amps, and the voltage across the resistor will be 30 volts. If you add a second 10 Ohm resistor in series, the load on the power supply is now 20 Ohms, so, by Ohm's Law, the current through the resistors will be 1.5 amps, and there will be 15 volts dropped across each resistor, for a total voltage drop of 30 volts.

If you put the two 10 Ohm resistors in parallel across the 30 volt supply, each resistor will now see 30 volts, and will each pass 3 Amps, so the supply will have to supply 6 Amps.