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.]
Best Answer
The free magma on $X$ consists of finite sequences of length $1$ or $2$, which consist of finite sequences of length $1$ or $2$, etc., of elements of $X$; the magma operation is just concatenation, i.e. $mn = (m,n)$. For example, $(a,((b,c),d))$ is such a sequence, where $a,b,c,d$ are in $X$. Elements may be visualized as finite binary trees, whose leaves are labelled in $X$. The examples gives:
To give a formal construction, define by recursion the sets $X_n$ of elements of height $n$ by $X_0 = X$ and $X_{n+1} = X_n + X_n^2$ (disjoint union). Then the disjoint union of $X_n$ is the free magma on $X$.
Now the free commutative magma is obtained by taking the quotient with respect to the smallest congruence relation satisfying $(a,b) \sim (b,a)$. This can be visualized with trees: Every branch can be rotated freely. So for example, now we don't distinguish between the trees
Note, however, that this does not justify that we may replace every bracket $(...)$ with $\{...\}$. Namely, $(a,a)$, which has two leaves, has to be distinguished from $(a)$, which has only one.
Remark: These constructions are "abstract nonsense", the same procedure works for free algebraic structures of any type (free monoids, free groups, free Lie algebras, etc.). Usually a bit of work has to be done to find normal forms for the elements of free algebraic structures. For free magmas, every element is already in normal form. For the free commutative magma on $X$, choose a total order on $X$, and call an element in normal form if the occuring elements of $X$ (ignoring the brackets) are ordered from left to the right. In the picture above, when we order $a<b<c<d$, then the tree on the left is the normal form of the tree on the right.