I'm thinking there are two ways the product forms. One is as Disco illustrated with the migration of the double bond and the other is to form the tri substituted double bond directly. I'm thinking if one took 2-methylbut-1-ene and treated it with acetic acid and potassium acetate, the double bond wouldn't migrate. If so, the the double bond could have been tri substituted from the start, no migration.
Just for the sake of argument, the mechanism could have acetate elimination to an alpha-beta unsaturated anhydride and then ketene formation. Since I argue Hoffman products look like removal of the most acidic protons, hence relatively poor leaving groups, ammonium salts and fluorides. Zaytzev products are more SN1-like, hence bromides and iodides in alcohols. The migration requires either protonation or base catalyzed migration. Neither seem likely, but that is just a guess. If elimination occurred first, this would increase the acidity of the delta hydrogen by being adjacent to an electron withdrawing sp2-carbon and in conjugation to the anhydride.
The reaction lacks detail, yields, by-products, etc. so, I am only guessing. I certainly don't know the mechanism.