For a finitely generated module $M$ over a commutative ring $R$, the first definition gives a rank function $r_M: \operatorname{Spec} R \rightarrow \mathbb{N}$, whereas the third definition gives $r_M((0))$.
When $M$ is projective, $r_M$ is locally constant, so when $\operatorname{Spec} R$ is connected -- so when $R$ is a domain -- $r_M$ is constant and may be identified with its value at $(0)$. I think you are asking for reassurance that for finitely generated non-projective modules over a domain, the rank function need not be constant. That's certainly true: take $\mathbb{Z}/p\mathbb{Z}$ over $\mathbb{Z}$ (or even over $\mathbb{Z}_p$): then the rank function is $1$ at $(p)$ and otherwise $0$, so the first definition really is not the same as the third.
Is this a problem? I don't think so. Asking for terminology in mathematics to be globally consistent seems to be asking for too much: even more basic and central terms like "ring" and "manifold" do not have completely consistent definitions across all the mathematical literature: rather, they overlap enough to carry a common idea. That is certainly the case here.
Similarly, asking which of 1) and 3) is "right" doesn't seem so fruitful. It is true that the first definition records more information than the third definition. But it's just terminology, and it is often useful to have a term which records exactly the information in the third definition: e.g. the "rank of an abelian group" is a very standard and useful notion, and usually often it means 3). (I am a number theorist, and in number theoretic contexts this definition of rank is quite standard. Apparently it is less so elsewhere...) For a finitely generated module over a PID, of course the rank in the sense of 3) is exactly what you need in addition to the torsion subgroup in order to reconstruct the module. In this context the rank function 1) gives some information about the torsion but not complete information -- e.g. $\mathbb{Z}/p\mathbb{Z}$ and $\mathbb{Z}/p^2\mathbb{Z}$ have the same rank function -- so the rank function as in 1) does not seem especially natural or useful. I am (almost) sure there are other contexts where it would be natural and useful to think in terms of the rank function as in 1).
Added: Following quid's comments I checked out Fuchs's text on infinite abelian groups, and indeed the rank of an arbitrary abelian group is defined in a way so as to take $p$-primary torsion into account. (I simply didn't know this was true.) Thus the rank defined there would be the sum over all the values of the rank function in the sense of 1).
All this seems to indicate, even more than my answer, that different notions of rank proliferate.
Added Later: Quid also suggests consideration of the quantity $\operatorname{mg}(M)$, the minimal number of generators of an $R$-module $M$: this is a cardinal invariant which is (clearly) finite if and only if $M$ is finitely generated. Let $R$ be a Dedekind domain with fraction field $K$. For a maximal ideal $\mathfrak{p}$ of $R$ and a finitely generated $R$-module $M$, let $M[\mathfrak{p}^{\infty}]$ be the submodule of $M$ consisting of elements annihilated by some power of $\mathfrak{p}$. Then $M[\mathfrak{p}^{\infty}]$ is a finitely generated torsion module over the DVR $R_{\mathfrak{p}}$ and is thus a direct sum of $\operatorname{mg}(M[\mathfrak{p}^{\infty}])$ copies of $R_{\mathfrak{p}}/\mathfrak{p}^k R_{\mathfrak{p}}$. Let us define $tr(M,\mathfrak{p})$ to be this number of copies. (When $M$ is infinitely generated I believe one should also count copies of $K_{\mathfrak{p}}/R_{\mathfrak{p}}$ in a certain sense in order to recover Fuchs's $p$-primary torsion rank in the $R = \mathbb{Z}$ case. Let me omit this for now.)
Now define
$R(M) = r_M((0)) + \sum_{\mathfrak{p} \in \operatorname{MaxSpec} R} tr(M,\mathfrak{p})$
When $R = \mathbb{Z}$ then $R(M) = \operatorname{mg}(M)$ is the "total rank" in Fuchs's sense. More generally $R(M) = \operatorname{mg}(M)$ when $R$ is a PID. However, I wanted to point out that in general the function $\operatorname{mg}$ behaves rather badly. This is discussed in $\S$ 6.5.3 of my commutative algebra notes. In particular, when $M$ is finitely generated projective, $\operatorname{mg}(M) \geq R(M) (= r_M((0))$ but can be larger. However, it is much more restricted than what I knew about before reading the comments on this question. In particular, it follows from the Forster-Swan Theorem that when $M$ is projective of rank $n$ then $\operatorname{mg}(M) \in \{n,n+1\}$. (In an earlier version of this answer I knew only that $\operatorname{mg}(M) \leq 2n-1$ and "guessed" that it could be that large over suitable Dedekind domains. Not a terrible guess, perhaps, but not the most educated one either...)
Best Answer
The dual module of a finitely generated module is reflexive, that is, $M^{**}=M$, and reflexives are awfully close to projectives. Specifically, if $R$ is a Noetherian domain, then a module is projective if $Ext^i(M,R)=0$ for all $i>0$, and its reflexive if $Ext^i(M,R)=0$ for $i=1,2$.
It is also worth noting that every reflexive is the dual of some module, specifically of $M^*$. Therefore, your question amounts to "for what rings is every reflexive module free?" In this light, its very similar to the question of when every projective module is free.
From the above Ext criterion, its clear that if the global dimension of $R$ is less than or equal to 2, that being reflexive is the same as being projective. I would go so far as to conjecture the converse is true: that if gldim of R is 3 or more, that there is a non-projective module which is reflexive (and hence it is non-free).
If this conjecture is true, then the answer to your question is "rings with global dimension 2 or less, such that every projective is free". Of course, its not immediately clear what these are, but its a start.