Why is the denominator in a conditional probability the probability of the conditioning event? Quite a simple or at least short question: Why is $ \frac{P(A \cap B)}{P(B)} $ divided by $ P(B) $ for the conditional probability?
$ P(A | B) = \frac{P(A \cap B)}{P(B)} $
Random image to visualize: 
I actually would like to make use out of the tree display which I can't grasp at all. Why would I divide by $ P(A+) $ when I had to move along this path?
Sorry for the mix-up of $A+$ and $ B $..


Intuitive, hand-wavey explanation that I feel is lacking here. (This argument works 100% fine in the case of uniform probability with a finite number of outcomes, such throwing a fair die, or picking cards from a well-shuffled deck, but to work more generally it would need some rephrasing.)
The first probability formula you learn is
$$
P(\text{Something}) = \frac{\text{Number of outcomes where Something happens}}{\text{Total number of relevant outcomes}}
$$
We are going to use this three times.
First off, the probability $P(B^+\mid A^+)$ is given by
$$
P(B^+\mid A^+) = \frac{\text{Number of outcomes where }B^+\text{ and } A^+\text{ happen}}{\text{Number of outcomes where }A^+\text{ happens}}
$$
Now we divide numerator and denominator by the total number of outcomes. This does not change the value of the fraction:
$$
= \frac{\text{Number of outcomes where }B^+\text{ and } A^+\text{ happen}/\text{Total number of outcomes}}{\text{Number of outcomes where }A^+\text{ happens}/\text{Total number of outcomes}}
$$
Applying our basic probability formula again, to numerator and denominator separately, we do indeed get
$$
= \frac{P(A^+\cap B^+)}{P(A^+)}
$$
