Suppose the 8 men are A B C D E F G and H
When you write $\binom{8}{2}$ , it means any 2 of these 8 men are selected, say, A and B.
But now when you considered the remaining 6 men (and remaining 2 women) in $\binom{8}{1}$ , how do you know that you're going to choose ONLY from among C to H? Instead of A and B, any other 2 men could have been chosen, we don't know who was chosen.
So this way, it's possible that when you selected 2 out of 8 men, you chose A and B, and then when you chose 1 out of 8, you chose A again <-- This should not happen, as you have to choose a person only once.
Similarly for the 5 women, we don't know which 3 were selected, so it's possible that you count the same woman twice.
That's why, total number of ways is = number of ways to form a team of 3 women and 3 men + number of ways to form a team of 4 women and 2 men
= $\binom{5}{3} \times \binom{8}{3} + \binom{5}{4} \times \binom{8}{2}$
= 560 + 140 = 700
