Why is the total probability theorem expressed in this way? The total probability theorem states the following:
Let $A_1$,...,$A_n$ be a partition of $\Omega$. For any event B,
$$Pr(B)=\sum_{i=1}^n Pr(A_i)Pr(B|A_i)$$
We know that $Pr(B|A_i)= \frac{Pr(B\cap A_i)}{Pr(A_i)}$, therefore the $Pr(A_i)Pr(B|A_i)$ can simplify to $Pr(B\cap A_i)$.
Couldn't the above theorem then be directly expressed as:
$$Pr(B)=\sum_{i=1}^n Pr(A_i\cap B)$$
If yes, then why use the above representation?
 A: Yes, the theorem of total probability can be expressed as
$$P(B) = \sum_n P(B\cap A_n)\tag{1}$$
(which form you seem to prefer) instead of the more usual
$$P(B) = \sum_n P(A_n)P(B\mid A_n)\tag{2}$$
but the whole point of expressing $P(B)$ in the form $(2)$ is to remind the reader that $P(B)$, the unconditional probability of $B$, is just a weighted sum of the conditional probabilities $P(B\mid A_n)$. The weights are, of course, just the $P(A_n)$ which sum to $1$.  Put another way, the unconditional probability $P(B)$ is the average value (or expected value) of the (conditional) probability of $B$ given that various (mutually exclusive) conditions (the events $A_n$) have occurred. From this we can deduce that the unconditional probability $P(B)$ is necessarily bounded above by $\max P(B\mid A_n)$ and bounded below by $\min P(B\mid A_n)$. So, $(2)$ does have its uses. In my opinion, it provides more insight into the issues than $(1)$.
A: I guess the main reason is the immediate applicability to
the elementary version of Bayes' Theorem, in which (i) sets $A_i,$ for $i = 1, \dots k$ form a partition of the sample space,
(ii) $P(A_i)$ and $P(B|A_i)$ are known, and (iii) $P(A_1|B)$ is to be computed. [One of the applications mentioned by @jbowman.]
Then
$$P(A_1|B) = \frac{P(BA_1)}{P(B)}
= \frac{P(A_1)P(B|A_i)}{\sum_{i=1}^k P(A_i)P(B|A_i)}.$$
