Perhaps the words in the picture are explained at greater length in the text, or perhaps actual numbers are given for the dissociation constants (https://www.quora.com/Why-is-o-flurophenol-is-more-acidic-than-p-flurophenol).
As it turns out, the dissociation constants are 8.7, 9.3. 9.9 and 10.0 for ortho, meta, para and unsubstituted phenol. Ortho-fluorophenol is the most acidic - perhaps in spite of the possible hydrogen bonding. The inductive effect of the fluoro group is surely greatest at the ortho position.
In the series of nitrophenols (Acidity order of nitrophenols), the para (p$K_a$ = 7.16) is slightly more acidic than the ortho (p$K_a$ = 7.2), so it could be said that the ortho substituted compound, which would be expected to be more acidic (if you assume that the resonance effects at ortho and para positions would be similar), but isn't, we have to look for some other explanation, and hydrogen bonding could stabilize the neutral molecule.
Well, even if the actual effect is stronger than 0.04 p$K_a$ units, the hydrogen bonding effect seems to be rather small, and is overcome in the fluorophenols by the $-I$ effect of fluorine. Just scratch out the words "H-bond decreases acidity". (Lightly.)