Why there is only 4.5V at the 5V output of my arduino board?
Now that you've posted the schematic, a straight answer is possible.  That schematic is a sloppy mess, but I know that's not your fault.  It doesn't say much about the folks that designed your arduino board though.  Yucc.  It makes you wonder what else they didn't pay much attention to.

The board can be powered from USB or a external supply.  When from USB only, there is a diode from the USB power to the "5V" supply.  The USB supply voltage is nominal 5V, but can be lower than what you actually measured in some cases.  What you measured is about expected with 5V from the USB then with a diode drop in series.

It looks like there is at least a reverse blocking diode (D1) at the external DC power input.  That means if you hook up the external supply backwards the board won't run but nothing will be damaged.  You therefore didn't break anything, at least due to providing the wrong external supply polarity.