I'll consider the interval $[0,2\pi]$ for notational simplicity. Consider the matrix
$$
A = \left(
\begin{array}{cc}
\sin ^2\tfrac{\theta }{2} & - \sin
\tfrac{\theta }{2} \cos
\tfrac{\theta }{2} \\
-\sin \tfrac{\theta }{2}\cos
\tfrac{\theta }{2} & \cos
^2\tfrac{\theta }{2}
\end{array}
\right),
$$
which defines an $R$-linear map $p:R^2\to R^2$. Computing $A^2$ we see that $p^2=p$, so $p$ is idempotent, and its kernel is a projective $R$-module $P$.
Now consider the map $$
\phi : f\in M \mapsto(f(\theta)\cos \tfrac{\theta }{2},f(\theta)\sin
\tfrac{\theta }{2}) \in R^2.
$$
It is clearly an $R$-linear injective map, whose image is precisely the kernel $P$ of $p$. It follows that $M\cong P$, and this shows projectivity.
Non-freeness is more subtle...
There is a morphism of rings $\varepsilon:R\to\mathbb R$ given by evaluation at $0$. One can see that $P\otimes_R\mathbb R$ is of dimension $1$ over $\mathbb R$, so that if $P$ is free, then it is free of rank $1$. In that case, $M$ would be free of rank $1$: suppose so, and let $h\in M$ be a generator. It is immediate then that every element of $M$ has to vanish where $h$ vanishes. But one can easily find an element of $M$ whose only zero is not a zero of $h$.