When computing the odds of making "One Pair", we compute the odds of the player making a pair at all and subtract out the odds that they will be *forced* to make a higher hand (full house, trips, two pair), given that they have made the one pair. This is important when trying to figure out the probability that you will not foul your hand. With a pair of aces in the back and a pair of deuces up top, you care about your odds of making *exactly* a one pair hand in the middle (i.e., not "at least" one pair). With other hands to dump into, you can dump off your higher hands if necessary and retain just the one pair.
In the screenshot, the player has already made one pair. Therefore, the probability of making *at least* one pair is 100%. However, there is a small, but non-zero, probability that they will be forced to make a higher hand. In this case it would require hitting 4 of the 8 remaining tens, jacks, and queens (in the 43 card deck). As you can imagine, drawing perfect-perfect-perfect-perfect with only 8 counterfeit outs is a rare occurrence (in fact, it's about 0.06%) -- hence the 99.94% chance of making one pair. The Two Pair only requires hitting at least one of the 6 remaining tens or queens in four draws and the trips requires hitting at least one of the jacks.
The cases where this user has seen the odds be cumulative are probably instances where the other rows were fully populated. In this scenario, there is nowhere to "dump", so whatever you're going to make is what you're going to make.
