It is impossible to produce an example of a finitely generated flat $R$-module that is not projective when $R$ is an integral domain. See: Cartier, "Questions de rationalité des diviseurs en géométrie algébrique," here, Appendice, Lemme 5, p. 249. Also see Bourbaki Algèbre Chapitre X (Algèbre Homologique, "AH") X.169 Exercise Sect. 1, No. 13. I also sketch an alternate proof that there are no such examples for $R$ an integral domain below.
Observe that, for finitely generated $R$-modules $M$, being locally free in the weaker sense is equivalent to being flat [Bourbaki, AC II.3.4 Pr. 15, combined with AH X.169 Exercise Sect. 1, No. 14(c).]. ($R$ doesn't have to be noetherian for this, though many books seem to assume it.)
There's a concrete way to interpret projectivity for finitely generated flat modules. We begin by translating Bourbaki's criterion into the language of invariant factors. For any finitely generated flat $R$-module $M$ and any nonnegative integer $n$, the $n$-th invariant factor $I_n(M)$ is the annihilator of the $n$-th exterior power of $M$.
Lemma. (Bourbaki's criterion) A finitely generated flat $R$-module $M$ is projective if and only if, for any nonnegative integer $n$, the set $V(I_n(M))$ is open in $\mathrm{Spec}(R)$.
This openness translates to finite generation.
Proposition. If $M$ is a finitely generated flat $R$-module, then $M$ is projective iff its invariant factors are finitely generated.
Corollary. The following conditions are equivalent for a ring $R$: (1) Every flat cyclic $R$-module is projective. (2) Every finitely generated flat $R$-module is projective.
Corollary. Over an integral domain $R$, every finitely generated flat $R$-module is projective.
Corollary. A flat ideal $I$ of $R$ is projective iff its annihilator is finitely generated.
Example. Let me try to give an example of a principal ideal of a ring $R$ that is locally free in the weak sense but not projective. Of course my point is not the nature of this counterexample itself, but rather the way in which one uses the criteria above to produce it.
Let $S:=\bigoplus_{n=1}^{\infty}\mathbf{F}_2$, and let $R=\mathbf{Z}[S]$. (The elements of $R$ are thus expressions $\ell+s$, where $\ell\in\mathbf{Z}$ and $s=(s_1,s_2,\dots)$ of elements of $\mathbf{F}_2$ that eventually stabilize at $0$.) Consider the ideal $I=(2+0)$.
I first claim that for any prime ideal $\mathfrak{p}\in\mathrm{Spec}(R)$, the $R_{\mathfrak{p}}$-module $I_{\mathfrak{p}}$ is free of rank $0$ or $1$. There are three cases: (1) If $x\notin\mathfrak{p}$, then $I_{\mathfrak{p}}=R_{\mathfrak{p}}$. (2) If $x\in\mathfrak{p}$ and $\mathfrak{p}$ does not contain $S$, then $I_{\mathfrak{p}}=0$. (3) Finally, if both $x\in\mathfrak{p}$ and $S\subset\mathfrak{p}$, then $I_{\mathfrak{p}}$ is a principal ideal of $R_{\mathfrak{p}}$ with trivial annihilator.
It remains to show that $I$ is not projective as an $R$-module. But its annihilator is $S$, which is not finitely generated over $R$.
[This answer was reorganized on the recommendation of Pete Clark.]
Hi Pete, this sounds like a lot of fun! I wish I could be there (-:
Here is a concrete and useful property of flatness, you can explain it without using Tor. Suppose $R\to S$ is a flat extension.
Then if $I$ is an ideal of $R$, tensoring the exact sequence:
$$ 0 \to I \to R \to R/I \to 0$$
with $S$ gives that $I\otimes_RS = IS$. The left hand side is somewhat abstract object, but the right hand side is very concrete.
There are very natural extensions which are flat but not projective. For example, if $R$ is Noetherian and $\dim R>0$, then $S=R[[X]]$ is flat but never projective over $R$.
Best Answer
There are two questions here. First of all, yes, the argument is fine; secondly, yes, there are rings with finitely many minimal primes, but infinitely many associated primes. So all together, the criterion is slightly more general than the one by Raynaud-Gruson mentioned in the question, but the proof is much easier. It also admits a further generalisation (very slightly) as shown below.
First, the construction of a ring with finitely many minimal primes but infinitely many associated primes. It is obviously inspired by the ring $k[t,x]/(t^2,tx)$ which corresponds to the line with an embedded point in the origin, just that we take infinitely many embedded points. Of course, we have to leave world of polynomials for that.
Let $H$ be the ring of holomorphic functions on the complex line $\mathbb{C}$ (with coordinate $z$) and $f\in H$ a non-trivial holomorphic function with infinitely many zeros. Then $A = H[t]/(t^2,tf)$ is such an example. The nilradical of $A$ is the prime ideal $(t)$. Thus, $(t)$ is the unique minimal prime. However, for each root $\alpha\in\mathbb{C}$ of $f$, we easily see that $(t,z-\alpha) = \mathrm{ann}_{A}(t(z-\alpha)^{-1}f)$ is an associated prime of $A$. By assumption, there are infinitely many roots, hence infinitely many associated primes.
To give a complement to the arguments given in the question and the comments on MSE, here is the generalisation to schemes (using the same arguments).
This is a local question and we assumed the connected components to be open, so we may just assume that $X$ is connected and show that the rank is constant then. Let $F$ be a quasi-coherent flat $\mathcal{O}_X$-module with $F_x$ finitely generated for each $x\in X$. Let $\eta_1,\eta_2,\dots \eta_s\in X$ be the generic points of the maximal irreducible components $X_i := \{\eta_i\}^{-}$, $i = 1,2,\dots s$. We first have to show that the rank of $F$ is constant on each $X_i$. The basic thing to note is: if $y\leadsto x$ is a specialisation, i.e., $x\in\{y\}^{-}$, then $\mathcal{O}_{X,y}$ is a localisation of $\mathcal{O}_{X,x}$ and with respect to this structure we have $F_{y} = F_{x}\otimes_{\mathcal{O}_{X,x}}\mathcal{O}_{X,y}$. Therefore, if $F_x$ is a free $\mathcal{O}_{X,x}$-module, then $F_y$ is a free $\mathcal{O}_{X,y}$-module of the same rank. By definition, each $x\in X_i$ is a specialisation of $\eta_i$ and so the rank is constant on each maximal irreducible component.
To show that the rank is constant on our connected scheme $X$ we consider the subsets $V_k := \{x\in X\mid \mathrm{rk}_{\mathcal{O}_{X,x}}(F_x) = k\}$ disjointly covering $X$ as $k\geq 0$. Since the rank is constant on maximal irreducible components, each $V_k$ is a union of such. But there are only finitely many maximal irreducible components, so there are only finitely many nonempty $V_k$ and all of them are closed, hence, also open. Since $X$ is connected, the only possibility for this to happen is when $X = V_k$ for some $k\geq 0$, i.e., if the rank is $k$ at every point.