What you are missing is that current flowing in one place can cause a different amount of current to flow elsewhere. What is really being conserved in this case is power, not current. You could just as well ask that since 220 V is going in and 12 V going out, where do the remaining 208 V go? That question is basically identical to the one you asked, just with voltage and current flipped around, and just as invalid.
The total power out has to equal the total power in, at least in the long term. Power is voltage times current. Either voltage or current by itself won't give you a useful picture. In this case, we have up to 200 mA in at 240 V, which is 48 W. What comes out is 12 V at 5 A, which is 60 W. Since this can't be, the specs above are inconsitant with themselves. Someone is either mistaken, being rather loose with the figures, or outright lying.
Lets work backwards to see what the input current really needs to be. 12 V at 5 A is very clearly 60 W out, so at least 60 W have to go in somehow. In reality, these things aren't 100% efficient, so a bit more than 60 W has to go in. The remaining power that doesn't come out electrically goes to heating the unit.
Let's say this power supply is 90% efficient, which isn't a bad number at all. In the 80-90% range is common. Something like 94% would be quite impressive. 60W / 90% = 67W, which is the power that must go in. 67W / 220V = 303mA, which is the input current required at 220 V due to conservation of energy alone.
Something is clearly wrong with your figures.
Once it became clear your figures were impossible, we got sidetracked by that and didn't get into much detail on your original question.
As I said briefly above, current in one place can cause more current in another place, even though the two current flows are not connected (totally different piles of Coulombs being moved around).
Another point you are missing is that the input current flows both into and back out of the device, and so does the output current. This power supply has basically 2 pairs of terminals, one for the input and one for the output. The 300 mA flowing into one of the input terminals comes back out the other input terminal. The same is true for the output current. The output current goes out the + terminal, thru your device that it is powering, and back in thru the - terminal. In the full power case, you have both 300 mA going in and out one pair of connections, and 5 A going in and out the other pair of connections.