How can a PSU be capable of providing more current than it draws from the mains
What power supplies do is a conversion of power, not current.

This means you take primary power (\$P_{pri} = V_{pri} I_{pri}\$) from the mains network and you get secondary power (\$P_{sec} = V_{sec} I_{sec}\$) at the output. The power at the output can never be greater than the power at the input:

$$P_{pri} &gt; P_{sec}$$

Substituting the above equations, we get:

$$V_{pri} I_{pri} &gt; V_{sec} I_{sec}$$

Let's solve for \$I_{sec}\$:

$$I_{sec} &lt; \frac{V_{pri}}{V_{sec}} I_{pri}$$

You see that the "trick" is hidden within the relationship of the input and output voltages. If the output voltage is lower than the input voltage, the output current may well be higher than the input current.

With the numbers you give (and assuming you live in the UK because this is where fuses are often found in mains plugs):

$$I_{sec} &lt; \frac{240V}{5V} 13A$$

$$I_{sec} &lt; 624A$$

This is just meant to prove the point. The number is very (!) much theoretical and way beyond the real numbers, because (i) the fuse has a rating much higher than the RMS current used by your power supply, usually around a factor of 5 to 10, because it must not blow during inrush events at turn-on, (ii) the power supply's output current may be limited by other factors and (iii) the power supply has some losses.

This relationship can also be explained using the waterfall model, where voltage equals height and current equals the size of the stream. On the input side, you can imagine having a very high waterfall with not that much water, on the output you have a big river with a lot of water but not much height that pushes the water towards the ocean.