An 82 ohm resistor would dissipate about 175 watts; indeed, a 15 watt bulb may dissipate that much power briefly. Something the size of a 15-watt bulb that dissipates 175 watts is going to get very hot, however, and when tungsten gets hot its resistance increases by an order of magnitude. You may have been expecting your 1K resistor to dissipate about 1.5 watts, but in practice it would have dissipated about 17.5 initially. Most resistors don't nicely increase their resistance with temperature the way a lightbulb filament does, however, so your resistor may have continued to dissipate that much power until it melted, whereupon its resistance might have actually decreased, causing it to try to dissipate even more power, until things went downhill very severely.
I would suggest that you might consider a Christmas decoration that's designed for a C7 bulb; such bulbs draw about 1 Watt. If you wanted to use a resistor that drew less power, two 33K resistors half-watt resistors in series would probably be reasonably safe if they were somewhat ventilated (the resistors should dissipate less than 1/8 watt each, but half-watt resistors should be used so that in case one fails shorted the other would be able to handle the 0.44-watt fault power) but might not let through enough power to operate the touch switch (if it does work, keep adding 33K resistors until it doesn't work reliably, and then remove one). The fact that incandescent bulbs have a much lower resistance when cold than when heated allows a much smaller energy-wasting element to be used for a given application than would otherwise be necessary.
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.
Best Answer
The cold resistance of your lamp is only around 1 Ohm. As such most of the voltage is being dropped over the resistor.
simulate this circuit – Schematic created using CircuitLab
If you want to drive the lamp from 5v you need to measure the hot current through the lamp at 1.5V so you can calculate the hot resistance. Knowing that you can select the right size of resistor. Be aware it will be small and will need to be high wattage.