I have a string of led lights that runs on two 3V batteries. When I measure the voltage on my multimeter it comes to about 6V as expected. I would like to use an adapter to avoid using batteries, so I took a USB phone charger and exposed the two voltage wires. The voltage output from the adapter is exactly 5V. When I connect the adapter to the lights, the lights get hot and shine too bright as if they have too many volts going through them. But there is only 5V as compared to the 6v coming from the batteries. Why is it that the adapter is making the lights do this? Thanks.
If these "3 volt" batteries are little coin cells, they have high internal resistance (at least 10 ohms, perhaps much more):
That high internal resistance acts as a current limiting resistor. If you have no other resistor on your LEDs, then you absolutely must add one to use with a 5v supply or you will burn up the diodes.
As you say in a comment, the batteries under load sit at 3.5 V. A properly regulated 5 V power supply that can deliver enough current sits at 5 V, load or no load.
So, more current passes through the LEDs with the 5 V power supply attached than with the batteries attached.
The open voltage of the batteries is irrelevant here; what matters is what the batteries do in a circuit where current flows. The batteries' internal resistance that makes their voltage drop under load is limiting the current that flows in the circuit.
Using a 6 V battery with a very low internal resistance, like a big lead-acid battery, would make a larger current flow than your 5 V power supply, because it would stay at about 6 V under load.