1) It is not true in  the second definition that for an arbitrary ring a) is equivalent to c).  
Indeed Bourbaki in  *Algèbre commutative, Chapitre II, Exercices §5, 12) c)*  exhibits a ring $B$ and a projective module of rank $1$ over $B$ which is not isomorphic to an invertible fractional ideal of $B$. This does not contradict Eisenbud's Theorem 11.6 because $R$ is explicitly supposed noetherian there.   
  
2) I think the only reasonable definition of $\operatorname {Pic(R)}$ valid for any commutative ring is to define it as the Picard group of the affine scheme $X=\operatorname {Spec}(R)$.  
As is the case for any locally ringed space $(X,\mathcal O_X)$ the Picard group consists of isomorphism classes of locally free $\mathcal O_X$-Modules of rank one.  
This is exactly the definition used with much success for general, non-affine, schemes but also for topological spaces, differential manifolds, etc.  

3) In our special case $X=\operatorname {Spec}(R)$ the definition in 2) translates in purely  algebraic terms to:   
$\operatorname {Pic(R)}$ consists of isomorphism classes of $R$-modules $M$ such that there exist finitely many elements $r_1,\cdots,r_n\in R$ with:  
$\alpha$)  $\sum Rr_i=R$    
$\beta$)   $M_{f_i}$ is a free $R_{f_i}$-modules of rank $1$ for all $i$.  
(Remark: $\beta$) implies that $M$ is finitely generated over $R$)  

4) The modules $M$ in  3) can also be characterized as the finitely generated  modules over $R$ such that  equivalently:  
i) $M$ is projective of rank $1$   
ii) The modules $M_\frak m$ are free of rank $1$ over $R_\frak m$ for all maximal ideals $\mathfrak m\subset R$   
iii) The canonical $R$-linear map $M\otimes_RM^*\to R:m\otimes \phi\mapsto\phi(m)$ is bijective  
Note that these are pleasant algebraic characterizations of the modules generating $\operatorname {Pic(R)}$, but the better, conceptual definition is that given in 2) and 3).  
**A final caveat:**   
The implication ii)$\implies$ i) above is false if one does not assume that $M$ is finitely generated over $R$.  
A counterexample is to be found in Bourbaki *Algèbre commutative, Chapitre II, Exercices §5, 7) b)*