For the half reaction HSO3- -> S2O6^2- why is it that 2e- are on the right hand side when balancing and not one electron?
Why is $2$ placed in front of the $\ce{HSO3}$? 


This is, of course, the elemental law of the conservation of mass. You have $x$ atoms that you put into your equation, you need to get $x$ atoms out again. While hydrogen and oxygen can be taken care of with water, you need to account for the sulphur immediately. Hence you need two sulphurs.


  The oxidation number of S on the left is +IV and +V so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one?


The oxidation numbers are, of course, correct, but remember that in the first step you wrote a $2$ before the sulphur. Therefore you now have two sulphurs that you need to treat with the appropriate number of electrons — either by stealing or by gifting.

Since you have two sulphurs and they each lose one electron, you need to add $\ce{2 e-}$ to the right.



Note: The OP already understood what went wrong through hints in the comments. Unfortunately, OP decided not to write up a self-answer.