OP: [...] *counterexamples can illuminate a definition (e.g. a projective module that is not free)*, [...]

Indeed, let our ring $\ \mathcal R\ $ be the the ring of all continuous functions from the Euclidean sphere $\ \mathbb S\ :=\ \mathbb S^2\ $ (or more generally,
$\ \mathbb S\ :=\ \mathbb S^{2\cdot n},\ $ where $\ n\in\mathbb N).\ $ Then module $\ \mathbb T\ $ of all continuous vector fields that are tangent to
$\ \mathbb S\ $ is a direct summand of free module $\ \mathcal R^3\ $ hence $\ \mathbb T\ $ is projective but it is not free.

The last property of $\ \mathbb T\ $ that states that $\ \mathbb T\ $ is not free is implied by the Karol Borsuk's theorem about the unruly hair on sphere $\ \mathbb S\ $ that is impossible to brush smoothly.