Well, your module M is certainly flat, because flatness is a local property (Atiyah-McDonald, Proposition 3.10, but it is also basically easy; first show being 0 is a local property, then show injectivity of morphisms is a local property, now show flatness is).
Note that this implies that any finitely presented such module is projective, since a finitely presented flat module is projective.
I suspect there is a counterexample in general, even with a finitely generated flat module. Try A a commutative von Neumann regular ring, like an infinite product of fields
$A = k_1 \times k_2 \times ... $
Take a finitely generated module that is not projective (but is necessarily flat, since the ring is von Neumann regular), which must exist but which I am having a little trouble writing down at the moment. I bet this will do the job. If no one else fixes this, I will try to do so later.
EDIT: A few minutes thought made this clear. Take I to be the direct sum of all the copies of k_i, and M to be the quotient A/I.
The (prime =) maximal ideals of A are the ideals m_i consisting of the elements which are 0 at every coordinate except the ith one. The local ring is k_i, which is a field. So actually any module M is locally free, but this particular M is not projective (because then the direct sum of the k_i would have to split off the product as a summand).
RE-EDIT: OK, so there are more maximal ideals than the m_i. However, for any commutative ring R, R is von Neumann regular if and only if the localizations of R at its maximal ideals are fields (Lam, A first course in noncommutative rings Ex. 4.15). Thus, any module M is max-locally free (and prime-locally free because primes =maximals for vNR rings). So any non-projective module, such as the one above, is a prime-locally free module that is not projective.
However, I bet think this module is probably not Zariski-locally freeas well, but I admit I am . So it shows that (3) does not sure imply (1), but says nothing about that. (4) implying (1).