There's more than enough information: the answer is that $S$ is *never* projective when it isn't "obviously" projective (i.e., never happens when the leading coefficient of $f$ is a non-unit). This is an application of Zariski's Main Theorem (in EGA formulation). There is no need for hypotheses about non-divisibility, so let's forget about that.

An obvious necessary condition for $S$ to be $R$-flat is that the coefficients of $f$ have total gcd equal to 1 (indeed, otherwise $f$ is divisible by some nonzero non-unit $r \in R$, so $S$ would have nonzero $r$-torsion, contradicting $R$-flatness of $S$). But this condition is also sufficient. Indeed, to check sufficiency we may assume $R$ is local (that step preserves the gcd hypothesis), and then some coefficient of $f$ must be a unit in $R$. Hence, $f_0 := f \bmod \mathfrak{m}_R$ is nonzero in $k[T]$ with $k = R/\mathfrak{m}_R$, so $f_0$-multiplication on $k[T]$ is injective and thus $f$-multiplication on the $R$-flat $R[T]$ has $R$-flat cokernel $S$ by the usual flatness criteria (see Cor. to Thm. 22.6 in Matsumura's "Commutative Ring Theory").

That being said, now let's assume (as we have seen we must) that the coefficients of $f$ have total gcd equal to 1 in $R$, so $S$ is a *quasi-finite* flat $R$-algebra. I claim that $S$ is projective if and only if the leading coefficient of $f$ is a unit in $R$ (in which case clearly it is finite flat and finitely presented as an $R$-module). The "if" direction is obvious. Now suppose the leading coefficient is not a unit, so $R$ is not a field and
we aim to show that $S$ is *not* projective as an $R$-module.
Note that if $S$ were projective as an $R$-module then the same would hold after localizing at any prime, so it is harmless to localize at a height-1 prime containing the non-unit leading coefficient of $f$. We assume $S$ is $R$-projective and seek a contradiction.

Now $R$ is a dvr and some lower-degree coefficient is a unit. It is likewise harmless to extend scalars to the completion of $R$ so that $R$ is complete. But now we can apply Zariski's Main Theorem in the EGA formulation to the quasi-finite $R$-algebra $S$ to get an $R$-algebra decomposition $S = S_{\rm{fin}} \times S'$ where $S_{\rm{fin}}$ is *finite* over $R$ and $S'$ has empty special fiber. If $S'$ is nonzero then it is a nonzero finite-dimensional vector space over the fraction field $K$ of $R$, so then $K$ would occur as an $R$-module direct summand of $S$. But $S$ is $R$-projective, so $K$ is $R$-projective, an absurdity (as $K$ is divisible, unlike any projective $R$-module). Thus, $S'=0$, so $S$ is $R$-finite, and hence by projectivity it is a free $R$-module. But the generic and special fibers of $S$ over $R$ have unequal ranks since the leading coefficient of $f$ is a non-unit and some lower-degree coefficient is a unit, contradicting $R$-freeness. Thus, $S$ wasn't $R$-projective after all.

QED

Everyquasi-finite flat algebra $S$ overanynoetherian ring $R$ that is not $R$-finite is not $R$-projective. Indeed, a separated quasi-finite flat map between noetherian schemes is finite if and only if its fiber rank is locally constant (proved in Deligne-Rapoport via Zariski's Main Theorem), so the fiber-rank of $S$ is non-constant on some connected component of Spec($R$). We can then make base change to a suitable complete dvr to reduce to the case when $R$ is a complete dvr and the special and generic fibers have different rank. Then the argument with ZMT in the answer applies. $\endgroup$ – grghxy Aug 17 '15 at 4:49