The battery voltage changing from 10.9 to 14.8 is no problem at all. Not for the 7805 and not for the transistors and not for the relays.
But I would add a fuse. If your circuit accidently causes a shortcut, you don't want the wires to become hot glowing.
The TIP102 can do 8A continuously, and 15A peak according to the datasheet.
It is a darlington transistor, so it needs very little base current (good), but has a larger voltage drop (not a big problem).
You have to calculate the base resistor.
Suppose the load is 200mA. At 200mA the hFE is 900 (datasheet).
200mA / 900 = 0.22mA
For saturation 30% is added : 0.22mA * 1.30 = 0.29mA
The VBE could be 2.8V
So the resistor is (5 - 2.8 ) / 0.29mA = 7k6
Calculation (copy whole link) : http://www.google.com/search?q=%285-2.8%29%2F%280.2%2F900
But since the base current is allowed to be 1A, I would choose a resistor of 4k7 (or even 2k2).
Any "logic level" mosfet that can handle the current will do.
You can search Ebay for : logic mosfet
I have a bunch of RFP12N10L for this kind of loads, because they were cheap at that moment.
It is best to have a resistor from the Arduino to the gate of 1k for the capacitance of the mosfet.
And a resistor of 10k from the Arduino to ground, to keep it low during power up.
Using a transistor or darlington transistor is simpler, so I advise to start with that TIP102.
You could also use the TIP120, it's about the same as the TIP102 : http://www.arduino.cc/cgi-bin/yabb2/YaBB.pl?num=1218993473
I have checked the datasheet, and the base resistor calculation turnes out the be almost the same.
If you use a transistor or a mosfet, you need a flyback diode over the relay.