What is the intuition behind the product rule of probability? Let's say that I have two independent events $A$ & $B$ from two different experiments. 
Assume that $P(A) = 3/13$ i.e. Out of the total 13 ways that an experiment can be performed, 3 ways make $A$ happen.
Assume that $P(B) = 7/13$ i.e. Out of the total 13 ways that another experiment can be performed, 7 ways yield $B$.
The product rule says $P(A \cap B) = P(A) * P(B)$. In this case the combined probability is $$P(A \cap B) = P(A) * P(B) = \frac {3}{13} * \frac {7}{13}$$ This makes sense to me that $13 * 13$ is actually the total number of ways that both experiments can be performed together & $3*7$ is the number of ways that $A$ and $B$ together can happen (by the fundamental rule of counting).
But this intuition quickly falls apart if $A$ & $B$ were dependent events of the same experiment. Let's say that experiment can be performed in 13 ways. $P(A) = \frac {3}{13}$ & $P(B) = \frac {7}{13}$ and lastly, $P(B|A) = \frac 13$. Now ,
$$P(A \cap B) = P(A) * P(B|A) =\frac {3}{13} * \frac {1}{3} = \frac {3}{39}$$
So $\frac {3}{39}$ shows 3 ways out of "39" ways the two events of the experiment happen. But the original experiment could only be done in "13" ways, what does 39 here represent?
I can understand if its unclear what I'm trying to convey. In essence, what is the intuition behind the product rule of probability? 
 A: Well, you've sort of found out that if your configuration space is $\Omega=\{1,2,...,n\},$ your probability distribution is 
uniform (say you're rolling with a fair $n$-sided die as opposed to considering the sum of eyes on two fair six-sided dies, for instance), then any event $A$ such that $\mathbb{P}(A)=\frac{k}{n}$ with $k$ and $n$ co-prime can never be independent of any other non-trivial event - if $\mathbb{P}(A\cap B)=\mathbb{P}(A)\mathbb{P}(B)=\frac{kj}{n^2},$  then there needs to be a set in $\Omega$ with probability $\frac{kj}{n^2}$, which can only happen in the above set-up if $n$ divides $kj$.
However, if $A$ and $B$ are events corresponding to different trials of experiments that are essentially the same, then you're back to the situation with a "natural" configuration space of cardinality $n^2$, and thus, you get independent events. This should be somewhat intuitively satisfying: It's not all that clear how many independent sub-events of a single experiment there really should be, unless the experiment involves independent components - say you're rolling two dice, or making measurements on two people and so on and so forth.
A: You can instead look at this from the perspective of $P(B|A)$. It represents the chance of $B$ occurring provided $A$ has occurred. The sample space can be thought of as having shrunk to $A$ and the event in question can only correspond to outcomes from $A\cap B$. However we cannot assign the probability $P(B|A)$ as $P(A\cap B)$ because the probability of the whole sample space must be $1$. A natural fix is to scale all the probabilities in the new sample space by dividing by $P(A)$ (provided it is positive). 
The definition of conditional probability follows. Hence, the multiplication in question becomes a scaling factor.
A: I don't think it falls apart. Consider this infamous example. 
Suppose a rare disease by the name conditionitis affects $1$% of the population. Let $D$ be the event that a person has the disease and let $T$ be the event that he tests positive.
Suppose that the test is $95$% accurate; there are many different measures of accuracy of the test, but in this example I will assume it to mean $P(T|D)=0.95=P(T^C|D^C)$. 
Let's ask ourselves, "What are chances that the test returns positive and the person has the disease?" We are interested in $P(TD)$. By the product rule - 
$$P(TD) = P(T|D)\cdot P(D)$$
That makes intuitive sense. Assume a sample of $10000$ people. $100$ people are likely to have the disease(the event $D)$ and $9900$ do not have the disease(the event $D^C$). Now, a $95$% accuracy means $95$ of $100$ suffering from conditionitis will test positive(the event $TD$). But, the catch is the test will falsely return positive for $0.05 \times 9900=495$ people. You don't wish to count these $495$ people.
So, when computing the joint probability $P(AB)$ in an experiment, you would like to count the occurrences of the event $A$ but restricting your attention to the trials where $B$ is observed (that is the event $A|B$), otherwise you risk counting $AB^c$.  
Hope that clarifies.
A: You can define indipendence in an other way
$$A \perp B \iff P(A|B)=P(A)$$
Which means that che event B doesn't affect A.
Now if you agree that the definition of
$$P(A|B)=P(A,B)/P(B)$$
Is intuitive, It becomes trivial that
$$A \perp B \iff P(A,B)=P(A)P(B)$$
