As far as I can think, the ortho effect comes into play which makes o-formyl benzoic acid more acidic.
But according to some, due to Hydrogen bonding between aldehyde and carboxylic group, hydrogen atom will be tightly bonded to it's two neighbouring oxygen atoms decreasing acidic character of o-formyl benzoic acid.
These two assertions are confusing and hence I seek for an experimentally verified result along with a satisfying explaination.
You can't compare them at all. If the ortho form dissociates at all then the carboxylate ion combines with the aldehyde function, forming a lactone ring. See Wikipedia.
This question invokes a comparison of the similarly structured nitrobenzoic acids. There the ortho compound is found to be strongest, beating its isomers by about one $pK_a$ unit and apparently indicating that the ortho effect is stronger than intramolecular hydrogen bonding.
Upon further review, we find that it wasn't a fair fight. Count the number of atoms that would be formed in a hydrogen-bonded ring: the hydroxyl portion of the carboxylic group, the carboxyl carbon, two aromatic ring carbons and two of the atoms from the nitro group. We generally do not get good intramolecular hydrogen bonding with a seven-atom ring, the arrangement just does not work sterically. Intramolecular hydrogen bonds are more common with five- or six-atom rings.
So in the case of 2-nitrobenzoic acid versus 4-nitrobenzoic acid the real answer is that hydrogen bonding is not in the game. Nor would it be for the formyl benzoic acids being described here, if the ortho compound were not prone to tautomerization; here too such a hydrogen bond would unfavorably form a seven-atom ring. We might say that the ortho compound, failing to form an intramolecular hydrogen bond, is thereby made prone to the strong dissociation and subsequent tautomerization noted above.
