Q:

What is the origin of the transferred oxygendianion in redox reactions? 

How come the oxygens transferred in redox reactions are always the $\ce{O^{2-}}$anion?  
For example, I have this set of rules, and the rules are implicitly referring to the $\ce{O^{2-}}$  anion, a potent base (otherwise the rules wouldn't make sense; for example, it makes sense that a base in acidic solution is protonated to water, and that a strong base in basic solution is leveled to hydroxide ion). 
In other words, how come, when electrons are transferred, such as in this unbalanced reaction: 
$\ce{ClO_{3}^- + 6I^- \leftrightharpoons 3I_2 + Cl^-}$
The $\ce{O^{2-}}$ anion is formed (if only to be consumed again)?  I remember my prof would say that specifically 3 $\ce{O^{2-}}$ anions "disappear".  
Why not the $\ce{O^{-}}$ anion or simply $\ce{O}$?  Does this have to do with stability?  

A:

I think you are misunderstanding the meaning of superoxide and the rules. 
Superoxide is $\ce{O_2^-}$
Superoxide is NOT $\ce{O^{2-}}$ as you have written. 
The rules are referring to hydroxide which is $\ce{OH^-}$, not superoxide. 
The reason hydroxide is used to balance equations in basic aqueous solution is:
$\ce{H_2O \leftrightharpoons H^+ + OH^-}$
There is hydroxide present in aqueous solution, but not superoxide, peroxide, hydroxyl radial, etc. 