Why is it that the transition function for DPDA's only works for 1 alphabet symbol, and 1 stack symbol?
Nondeterminism, means the machine has a choice. In both your examples, it indeed has a choice. 

Let's assume the stack has an 1 on its top, and the next input symbol is 0.
The PDA can choose whether to take the transition $f(q_i,1,0)\to(q_i,0)$ or  $f(q_i,\epsilon,0)\to(q_i,0)$. Note that the effect is different: in the first it deletes the 1 in the stack and puts a 0 instead of it. In the second it adds a 0 on top of the 1 without removing it. 

Therefore, there are two (different) paths the machine can go from this points, hence the nondeterminism.

Just to clarify: the definition for Deterministic-PDA is clear-cut: at any moment, and for any input/stack/state configuration there is only one possible transition. If this is not the case, the PDA is non-deterministic even if multiple transitions lead to the same configuration. It's just a matter of definition. 

However, in the question above, the two transitions may actually lead to different configurations, which makes the nondeterminism even more clear.