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 $..


The tickets-in-a-box model provides a helpful picture.  It makes the formula for conditional probability immediate and obvious.
Let $\mathcal{A}$ and $\mathcal{B}$ designate events: collections of types of tickets.  (That is, whenever a ticket for a particular outcome is in an event, all the tickets for that outcome must be in the event.)  Recall that the probability of $\mathcal A$ is defined as the proportion of tickets of type $\mathcal A$ in the box.  It is a ratio of that ticket count to the count of all tickets in the box $\Omega.$  (Some people insist all probabilities are conditional and to emphasize this would write the probability as $\Pr(\mathcal A\mid \Omega).$)
If we were to remove all tickets not of type $\mathcal{B},$ we would still have a box with tickets -- all of type $\mathcal B,$ obviously.  Assuming some tickets remain (that is, $\mathcal{B}$ is not empty), this is still a tickets-in-a-box model.
The selective removal can alter the relative proportions of $\mathcal A$ among the remaining tickets.  The new proportion of tickets of type $\mathcal{A}$ is their probability when $\mathcal B$ is viewed as the "box" -- the sample space.  It is called the conditional probability of $\mathcal{A},$ conditional on $\mathcal{B},$ and is written $\Pr{\left(\mathcal{A}\mid\mathcal{B}\right)}.$

In this Venn diagram, areas (relative to the total area of the rectangle $\Omega$ representing the whole probability space) represent probabilities.  It looks like both the events $\mathcal{A}$ and $\mathcal{B}$ have probabilities around $1/3$ each.  But when all non-$\mathcal{B}$ points are removed, the new sample space is $\mathcal{B}$ itself and the portion of $\mathcal{A}$ within it is tiny – just the sliver of space where the circles intersect, as a proportion of the area of the circle representing $\mathcal{B}.$  In this example $\Pr{\left(\mathcal{A}\right)}\approx1/3$ but $\Pr{\left(\mathcal{A}\mid\mathcal{B}\right)}\approx1/100,$ exemplifying how dramatically the conditioning on $\mathcal B$ can change the probabilities.
It should be apparent that (a) this construction works provided $\mathcal{B}$ does not have zero probability and (b) in that case, the new relative area of $\mathcal A$ is
$$Pr{\left(\mathcal{A}\mid\mathcal{B}\right)=}\frac{\Pr{\left(\mathcal{A}\cap\mathcal{B}\right\}}}{\Pr{\left(\mathcal{B}\right)}}.$$

The tree diagram is merely another way to (partially) represent the Venn diagram.  Branching at $\mathcal A$ corresponds to slicing everything in $\Omega$ into outcomes in $\mathcal A$ and outcomes not in $\mathcal A;$ likewise, branching at $\mathcal B$ slices everything by $\mathcal B.$
In your diagram you condition on $\mathcal A.$  This means removing everything from the tree that is outside $\mathcal A.$  Afterwards, only two leaves remain: everything in $\mathcal A$ and in $\mathcal B$ ($\mathcal A \cap \mathcal B$) and everything in $\mathcal A$ and not in $\mathcal B$ ($\mathcal A \cap (\Omega \setminus \mathcal B)$).  Those leaves represent all the tickets remaining in the box.  Trees are usually decorated with probabilities permitting you to compute the relative proportions of those leaves.
Translating between different ways of visualizing the same thing is a useful skill because some visualizations support specific concepts or calculations especially well, and others shine in other areas.
