I know that for an event $B$, we have Pr($B$) + Pr($B^c$) = 1. But when dealing with conditional probability, how am I to deal with this knowledge? For example, consider this example. Let event $A$ be the probability of coming first in a race, and let event $B$ be if it's raining outside.
It seems that it should be that
$$Pr(A=First Place|B=Raining) + Pr(A=Not First Place|B=Raining) = 1$$
My argument for this is that if both terms have the same condition, then you can really just ignore the condition. Why? Because instead, for example, let $C$ be if I am human, which is trivially true. Then
$$Pr(A=First Place) + Pr(A=Not First Place) $$
$$= Pr(A=First Place|C=Human) + Pr(A=Not First Place|C=Human) $$
$$= 1. $$
I can add on any other trivial condition and it will still be true, I argue. And therefore just consider if it's raining or not as a trivial condition.
The equation is true and your interpretation seems reasonable to me. If you want to confirm it algebraically, start with the definition: $$\eqalign{P(A|B)+P(A^c|B) &=\frac{P(A\cap B)}{P(B)}+\frac{P(A^c\cap B)}{P(B)}\cr &=\frac{P(A\cap B)+P(A^c\cap B)}{P(B)}\ .\cr}$$ Now the events $A\cap B$ and $A^c\cap B$ are mutually exclusive, so $$P(A\cap B)+P(A^c\cap B)=P((A\cap B)\cup(A^c\cap B))=P(B)$$ and the proof is complete.