Amortized value of dequeue with potential function
The intuition is following: the "bad" case when you have to do expensive operation is when one of the stacks is empty, and the other one is big. You need to compensate its real cost with a potential decrease.  Thus, you can define potential function as $2 \lvert \operatorname{size}(head) - \operatorname{size}(tail) \rvert$. 

(For simplicity I represent hard pop as combination of two operations - rearrangement of elements using the third stack and a light pop)

Before the rearrangement, one of the stacks is empty, and other one contains $k$ elements, and right after both contains $\frac{k}{2}$. Therefore, the potential decreases on $2k$ and amortized cost of rearrangement is

$(k + \frac{k}{2} + \frac{k}{2}) + (-2k) = 0 = O(1)$

The amortized costs of light pop and push are clearly $O(1)$ as well as potential increases on $2$ at most.