Naive Bayes Intuition
Because in general $p(v_i \mid a_1 \dots a_n) \ne \prod_j p(v_i | a_j)$.  There is no reason to expect that equation to hold with equality.  You can't just write down an equation and hope that it is true; you have to be able to justify it from the axioms of probability, and you can't justify the equation you want to hold.

Remember that $p(A \mid B)$ is not the same as $p(B \mid A)$.  The two probabilities can be very different.  Be careful not to confuse them, or you will definitely get very mixed up.

In general, if $A,B,C$ are events, then $p(A \mid B,C)$ is not necessarily equal to $p(A \mid B) \times p(A \mid C)$.  It is easy to come up with a counter-example.  For instance, suppose I toss a fair die, and define the following events: $A$ = "the die came up 5", $B$ = "the die is showing an odd number", $C$ = "the die is showing a prime number".  Then $p(A \mid B,C) = 1/2$, $p(A \mid B) = 1/3$, $p(A \mid C) = 1/3$, and $1/2 \ne 1/3 \times 1/3$.

The reason why you can write $p(a_1 \dots a_n \mid v_i) = \prod_j p(a_j \mid v_i)$ is because $a_1,a_2,\dots,a_n$ conditionally independent on $v_i$: once $v_i$ is fixed, then $a_1,a_2,\dots,a_n$ become independent.  In general, if $A,B$ are conditionally independent on $C$, then $p(A,B \mid C) = p(A \mid C) \times p(B \mid C)$ -- this is in fact the definition of conditional independence.  That's why $p(a_1 \dots a_n \mid v_i) = \prod_j p(a_j \mid v_i)$ is justified.  (However, no such justification is available for $p(v_i \mid a_1 \dots a_n) \ne \prod_j p(v_i | a_j)$.)

If you are not familiar with the concepts of independence and conditional independence, I suggest reviewing a standard probability textbook, where you can find more on that subject to help you practice and become proficient with those concepts.