Divisibility criteria of 24. Why is this?

The problem here might be something like $12$. You see, we have that $12$ is divisible by both $6$ and $4$, but it's not divisible by $24$. The reason they suggest $3$ and $8$ is because they are relatively prime, meaning that you can't have the sort of overlap in the case of $6$ and $4$.

This all has to do with the Fundamental Theorem of Arithmetic, which says that each number can be written uniquely as a product of primes, and primes have the special characteristic (or as Marvis points out, they are defined to be exactly those numbers with the characteristic) that if $p|ab$, then $p|a$ or $p|b$. So if $3$ and $8$ divide a number, then $24$ divides that number. But $6$ and $4$ dividing a number doesn't even guarantee that $8$ divides that number.

If $3 \mid n$ and $8 \mid n$, then clearly the $\mathrm{lcm}(3,8) = 24 \mid n$, since the least common multiple of $a$ and $b$ clearly divides any number divisible by both $a$ and $b$.

On the other hand, $4 \mid n$ and $6 \mid n$ is only enough to conclude that $\mathrm{lcm}(4,6) = 12 \mid n$.

If you are divisible by both $6$ and $4$, the number could be 12 which is not divisible by 24.

The reason for using $3$ and $8$ is that the least common multiple is 24. So every number that is divisible by both $3$ and $8$ is divisible by $24$.