Your battery solution will not work and may be dangerous.
You need something like this; (search for "dc dc converter"; anything car orientated will be suitable, given that car battery voltage is somewhere in the 12-15V range)
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 short answer is "don't do that."
The voltage dropped by a resistor is given by Ohm's Law: V = I R.
So if you know exactly how much current your device will draw, you could choose a resistor to drop exactly 7.5 V, and leave 4.5 V for your device, when that current is run through it. But if the current through your device is changing, or if you want to make more than one system and not every device is exactly alike in current draw, you can't consistently get 4.5 V at the device using just a resistor.
Your other options include
A linear regulator. This is basically a variable resistor that will adjust it's value to keep the output where you want it. This is probably only a good solution if your device draws very little power (maybe up to 100 mA).
A shunt regulator. This means using a resistor to drop the voltage like you are suggesting, but then adding an extra device in parallel with the load to control the voltage. The shunt regulator will adjust its current (within limits) to keep the current through the resistor correct to maintain the desired output voltage.
A switching regulator. This uses some tricks to generate your desired output voltage with much better power efficiency than a linear regulator. This is probably the best choice if your device needs more than 10 or 20 mA of current.