Specs on an AC-DC adapter that has two outputs

First, the over arching thing is that if the voltages match and the charger can supply more current the the original you were given, you are fine. If it supplies less, it may still be enough, especially if the battery is already charged.

1. Your wall socket will general be able to deliver in excess of 15A(depends on your country). The input current they expect is based on how much power their circuit actually uses.
2. Actually, their voltage is different, which is normally a larger issue for the adapters but the reason that interchanging is somewhat safe is for a hidden reason. Let me note first, plugging in a mismatched adapter can damage your device, and easily, if it is not able to handle the voltage you are going to supply. Most adapters user a communication protocol to verify the adapter was purchased from the original equipment manufacturer(OEM). If they do not detect this they often limit their power draw to "protect" the charger. This gives the advantage of protecting the charger if it is underrated.
3. Yes, you should be able to if the computer believes it is acceptable to use(no comm protocol to recognize a mismatched adapter). This voltage output is what primarily determines function. In electronics voltage controlled circuits are much easier to generate and use. This gets a bit too detailed, but I hope this information helps.

Here's what's going on.

The LM317 is what is known as a linear regulator. You are going from 18 volts to 6 volts. To do that with a linear regulator, the 12 volt difference is converted into heat. The regulator is trying to dissipate that, but it can't, so it's getting hot and going into thermal shutdown.

My guess is that the bench power supply is running a lower voltage than the laptop power supply

You have a few options.

You could reduce the input voltage to the regulator, to something like 9 volts. That would reduce the dissipation drastically, though it might not be enough. Note that linear regulators have what is called the "dropout voltage", which is a delta between the input and output voltage. The datasheet says it's about 3 volts, so 9 volts would be about the minimum.

You could add a heat-sink to the regulator; that would allow it to dissipate more heat. It would still get hot, but it would let the heat get away more easily.

You could use a different kind of power supply - known as a switch power supply (in this case, running in "buck" mode) that can convert from 18V to 6V at a much higher efficiency. They are relatively cheap.

Or, you could find a 6V power supply / adapter that will give you the current that you need. That would be the simplest to do.