Why $P(X|Y) = P(X_1,X_2|Y) = P(X_1|X_2,Y)P(X_2|Y)$? Let $X = <X_1,X_2>$, why
$$P(X|Y) = P(X_1,X_2|Y) = P(X_1|X_2,Y)P(X_2|Y)$$
What is the property of probabilities involved?
 A: So everything you've written has $Y$ as given. So we can just drop it and consider the simpler formula:
$$ P(X) = P(X_1, X_2) = P(X_1 \cap X_2) = P(X_1|X_2)P(X_2) $$
Just replace $P(\_)$ with $P(\_|Y)$ and the reasoning for why it works is the same. Now we just have to understand: 
$$ P(X_1 \cap X_2) = P(X_1|X_2)P(X_2)$$
This should be a little more intuitive, it just says that the probability that both $X_1$ and $X_2$ happen is the same as the probability that $X_2$ happens and also that $X_1$ happens given that $X_2$ has already happened. Alternatively, this is a rephrasing of Bayes theorem. 
A: Using the definition of conditional probability 
$$P(X_1, X_2|Y) = \frac{P(X_1, X_2, Y)}{P(Y)} = \frac{P(X_1|X_2,Y)P(X_2, Y)}{P(Y)} = \frac{P(X_1|X_2,Y)P(X_2|Y)P(Y)}{P(Y)} = P(X_1|X_2,Y)P(X_2|Y)$$
The second and third step in the equality comes from the general product rule of probability or chain rule. That is in general 
$$P(X_1,...,X_n) = P(X_1|X_2,...,X_n)P(X_2,...,X_n)$$
$$= P(X_1|X_2,...,X_n)P(X_2|X_3,...,X_n)P(X_3,...X_n)$$ $$= P(X_1|X_2,...X_n)P(X_2|X_3,...,X_n)P(X_3|X_4,...X_n)...P(X_n)$$
