If I remember correctly, there are at least three very recent/active threads from people who did not understand how voltage and current works. This is a recurring thing, and it is potentially dangerous. We should consider making a sticky.
Voltage and current are connected via Ohm's law: R = U/I.
R, the resistance, is in first approximation a fixed quantity of the circuit*. So if U, the voltage, is also given, then I, the current is fully determined. That is why a rating of 5V / 5A on a power supply means that it supplies 5V, and at these 5V, a MAXIMUM current of 5A can be provided. This would happen if a load of 1 Ohm would be connected. If someone would connect a load of 0.1 Ohm, the result depends on the power supply. The voltage could drop below 5V, the supply could become very hot or catch fire, a fuse could burn or it could just shut off.
Think of it like a water system:
- Voltage is equivalent to pressure. There can still be 2bar on a closed faucet and you can meassure this without spilling any water (e.g. by looking at how much a spring is pressed back by the pressure).
- Current is how much water flows through a pipe of a certain diameter at a given moment. This depends on the pressure of the faucet and the diameter of the pipe.
- Resistance can be understood as the diameter of the pipe
Now imagine a faucet and a hydrant. Both might be served by the public water system and have a given pressure, but the hydrant will be able to supply much more water at this pressure then your home faucet. However, if you connected a small enough pipe, both would send the same amount of water through and it would be fine - because they have the same pressure. If you connected a fire fighters hose to your home faucet, on the other hand, just what your faucet can supply would drip out of the hose with almost no pressure on the end.
So voltage (or pressure) is a static quantity. As long as you stay within the ratings, (again, first order) your voltage does not change, regardless of what you connect, be it a 20mA LED or a 2A tablet charger. The current, however, is a dynamic quantity that changes with the load, so you cannot meassure a single value for a supply. So what you did is, you actually shorted the batteries, which would in theory lead to infinite current. But since there is no perfected shortcircuit and "infinite current" is not in the specs of the batteries, what you actually did is you attached a load of the internal resistance of your meter + the wires + the internal resistance of battery, which results in a certain current. This current is not some principle property of your batteries, but rather determined by you setup. It is also not a safe current to use, since you noticed the the batteries became very hot.
No, I wrote "in first order" a few times. Let's have a quick look at the second order. The amount of water (=current) that your faucet can pump into your hose is determined by the diameter of that hose. But if the hose becomes larger, it is more and more determined by the diameter of the pipe going TO your faucet. This is the "internal resistance" of the supply. For some supplies, it will sufficiently limit the current. In your faucet, it usually does, that is why it is mostly safe to open a faucet with full power and nothing connected. It also means, though, that the pressure will not be constant, it will drop down the more water you take. At the hydrant, it will drop considerably later, because it is directly attached to the big pipes under the streat, with more big pipes to the hydrant.
In modern Electronics, on the other hand, the internal resistance is comparably much smaller. It is small enough that the maximal possible current is way higher than the parts of the supply can take. That is important to keep the voltage stable at all rated loads. So exeeding the current rating WILL damage a power supply that is not protected otherwise.
For batteries, internal resistance is usually not determined by wire resistance and such, but by ionic resistence. It is basically a virtual resistor that is determined by the speed at which the chemical reaction the battery can generate charge. So with batteries, you will often already see voltage drop in the range of the specs, but shorted out, they will usually still reach an amount of current that can be very dangerous and even cause the battery to explode. (On the other hand, this is why you always check the voltage of a battery WHILE it is driving a load, not with open contacts.)
You should also note that only because the current is save for the supply, and the voltage low and not considered dangerous, it can still do very bad and potentially dangerous things to the loads attached. A 4 Ohm 1/4W resistor (the common thing you got for your Arduno stuff) as a load on a phone battery will draw a current of 1A from the battery, which is totally fine, but the resistor will still dissipate 4W (P = U*I), which is 16 times it's rating and it will most probably catch fire. A carbon resistor (usually tan colored) will react a little safer that a metal film resistor (usually blue) in that regard. A 5V/1.2A phone charger with an unfortunately damaged plug may very well set your carpet on fire.
All of this was taught to you in high school physics, by the way, you should not have hated that subject and called it useless 
- It is, of course, not. The effective resistance of LEDs changes with the voltage, microcontroller draw different amounts of currents in different states and so on. But stay with me.
.