In Chain rule wiki, this is the first example (simplified):
User1 has 1 Black ball and 2 White balls and User2 has 1 Black ball and 3 White balls.
I guess this is the Joint table:
B W
U1:1 2
U2:1 3
Suppose we pick a user at random and then select a ball from him. Let event U1 be choosing the first user with P(U1)=1/2. Let event W be the chance we choose a white ball. The chance of choosing a white ball, given that we have chosen the first user is P(W|U1)=2/3. This intersection can be found by the chain rule:
P(W, U1) = P(W|U1)*P(U1) = 2/6
Question: From the table above, from my understanding, the P(U1) is 3/7 and not 1/2. Therefore, the chain rule would result in another Probability. What is the difference between these 2 different scenarios when chain rule is applied and how should I interpret this result?
Note: While yes, there are seven distinct possible outcomes for the scenario... they are not equally likely to occur! Your calculation leading to $\frac{2}{6}=\frac{1}{3}$ for $Pr(W,U_1)$ is correct. Your proposed calculation leading to $\frac{3}{7}$ for the same is not.
Recall, the probability of an event occurring is given by $Pr(A) = \frac{|A|}{|S|}$ where $S$ is the sample space is only a valid way of calculating the probability if it is known ahead of time that every outcome in the sample space is equally likely to have occurred. Remember, there are two outcomes to playing the lottery, you either win or lose... but winning doesn't occur half of the time.
If it is not clear why these outcomes are not equally likely to have occurred... consider the scenario where we have two users, like above, who we pick from with equal chance of $\frac{1}{2}$ (e.g. by flipping a fair coin). Suppose the first user has a single white ball and no others. Suppose the second user has a billion black balls and no others. After having selected our user, we pick a ball from him.
Clearly, half of the time we pick the first user and get a white ball while the other half of the time we pick the second user and get a black ball. The fact that there are so many more black balls than white balls is totally irrelevant here. Perhaps it would have been more relevant in the case that we had mixed all balls from all users and picked from that collection rather than first picking a user and picking balls from them, but that is explicitly not what the problem is getting at here.