Let A be a unital associative algebra over a field k.
Then A is smooth if and only if X:=Spec(A) is smooth. That is $\Omega_{X|Spec(k)}$ is locally-free. The later module is isomorphic to $\Omega_{A|k}$ the Khaler differentials on A over k.
Now a result equates local freeness with projectiveness of modules, therefore $\Omega_{A|k}$ is projective.
Now, the HKR theorem gives an isomorphism $\Omega_{A|k} \cong \Omega (A)$, where $\Omega (A)$ are the non-commutative (or abstract) differential forms on A.
Finally, a quasi-free algebra is one for which $\Omega$ is projective.
However, $k[X,Y]$ is smooth but can be seen to not be quasi-free, this leads me to believe there is a flaw somewhere in my reasoning, but I'm uncertain where?