I have mainly seen that the probability is 2.02%, but it always somewhere said it does not include if boards pairs. So I was thinking is the probability of getting 2 pairs if board pairs is equal to:
1 x (48/51) x (44/50) x (6/49) x (3/48) x 3 x 2 = 4.04%
44/50--> unpaired card
6/49 --> pairs one of the hole cards
3/48 --> pairs in one case hole card and in other the flop card
So I multiplied it by 2 because the equation for when the board pairs at least as I formed it was the same as when it pairs with hand.
Is this the correct reasoning or is there a mistake?
Your end result is correct, but your approach is not. Having unpaired cards is given, so don't take into account the probability of having unpaired cards.
To begin with, given that you hold two cards there are $\binom{50}{3}=19600$ possible three-card combinations for the flop. This is the denominator. For the numerator, you need to count the number of three-card combinations that give you two pair.
Without a board pair, you need a pair for your high card ($3$ choices), a pair for you low card ($3$ choices) and a third card of another rank ($44$ choices). Thus, there are $3*3*44=396$ three-card combinations for this case.
With a board pair, you have $11$ choices for the rank of the board pair, $\binom{4}{2}=6$ choices for the suits of the board pair, and $6$ choices for the third card to pair one of yours. Thus, there are $11*6*6=396$ three-card combinations for this case as well.
Adding these together gives you the total number of ways to hitting two pair. The probability of hitting two pair with unpaired hole cards is:
$$\frac{792}{19600}\approx0.040408$$