The problem is not that "there is not enough current", but that you can't dissipate enough power.
Let's assume you are powering the board from an external 12V supply. Now, the 5V are obtained from 12V with a resistive regulator: this means that the regulator (see this image if you are interested in knowing what component it is) simply behaves like a variable resistor, dissipating the power in excess. Now, let's assume you want to draw 4A from the 5V rail, since you want to turn on a lot of LEDs and other stuff. Since P=VI, the regulator (that very little component) has to dissipate (12V-5V)*4A = almost 30W. This means that it becomes a little heater and.. Puff! it breaks.
Now, 4A is a really high current, but that regulator can withstand very little power dissipations. According to the datasheet, and estimating a 1cm x 2cm heatsing under it, we can estimate a 25 K/W thermal resistance, which means that each watt you dissipate raises the internal temperature by 25K (=25°C). You can't go beyond 150°C without damaging the component, which means that if your room is at 25°C you can dissipate AT MOST 5W. In practice, never go beyond 3W. Anyway even 5W mean that, at 12V, you can't get more than 5W/7V=0.7A.
By the way, the regulator is rated 0.8A, so you should never go beyond that value in any case.
Now, this was the external regulator. As for the usb, you can't get more than 500mA from that power supply. That's written in the specs. For this reason, a 500mA fuse is on that rail, so going beyond 500mA will blow that one, interrupting your power. Well, that's a self-healing fuse, so after some time (maybe hours?) it will operate again, but.. Don't try that.
In the end, the digital pin current. The atmega datasheet says that
- The absolute maximum current per I/O Pin is 40.0mA
- The absolute maximum current inside VCC and GND Pins is 200.0mA
This means that from each single pin you can get at most 40mA (but I suggest you never go beyond 20mA), and the sum of all the currents (and the internal peripherals) should never exceed 200mA. So if you need to power 2 leds, you can give each of them 20mA, if you need to power 15 of them you can't (you are limited to, let's say, 10mA). If you need more, use transistors to separate the current paths.