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 $..


Sycorax has answered it nicely. But let me leave behind an intuitive set up that articulates the genesis of the concept.

Start with the very familiar experiment: throwing a die. Let $A:=\{\text{odd number appears}\}, ~ B:=\{\text{a digit smaller than or equal three appears}\}.$ One might wonder what the probability would be of $A$ had it already been assured that $B $ was realised.
Construct the probability space $(\Omega, \mathcal A, \mathbb P) ,$ with $\Omega= \{1, 2,3,4,5,6\}, ~\mathcal A= 2^\Omega$ and $\mathbb P$ being the discrete uniform distribution on $\Omega.$ Naturally $A=\{1, 3,5\}, ~B=\{1, 2,3\}.$ If $B$ has indeed occurred, then it is plausible to assume the uniform distribution on the remaining possible outcomes, i.e. $\{1,2,3\}.$ For assessing the new situation, define a new probability measure $\mathbb P_B$ on $(B, 2^B) $ by
$$ \mathbb P_B[C]:= \frac{\#C}{\#B}, ~C\subset B;$$ natural extension would be to assign probability $0$ on $\Omega\setminus B$ as
$$\mathbb P[C|B]:= \mathbb P_B[C\cap B]:= \frac{\#(C\cap B) }{\#B}, ~C\subset \Omega.$$
In that case $\mathbb P[A|B] =\frac{\# \{1,3\}}{\#\{1,2,3\}} = \frac23.$ This is different from $\mathbb P[A]=\frac12.$
This prompts us to define the conditional probability for any $B\in \mathcal A$ as a new probability measure  $\mathbb P[\cdot|B]$ on the space $(\Omega, \mathcal A)$ for $\mathbb P[B] > 0$ as
$$ \mathbb P[A|B] =\begin{cases}\frac{\mathbb P[A\cap B]}{\mathbb P[B]},~~ \mathbb P[B] > 0\\ 0 ,~~~~~~~~~~~\textrm{otherwise}\end{cases}.$$

Reference
Probability Theory: A Comprehensive Course, Achim Klenke, Springer, 2014.

After edit of OP:
As the definition developed above, that $A^+$ has occured provides new information has been incorporated by dividing by $\mathbb P[A^+].$
