1) About the second definition:  
$\alpha$) It is not true  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.  
  $\beta$) It is also not true that b) is equivalent to c):  
Take $R=\mathbb Z$ and $\mathbb Z\subsetneq M=\bigcup \frac {1}{p_1\cdots p_i}\mathbb Z\subsetneq \mathbb Q$ where $p_i$ is the $i$-th prime.  
 Then for all prime $\mathfrak p\subset \mathbb Z$ the $\mathbb Z_\mathfrak p$-module $M_\frak p$ is free of rank $1$, since $$M_{(0)}=\mathbb Z_{(0)}(=\mathbb Q) \quad  \operatorname {and}\quad M_{(p_i)}=\frac {1}{p_i}\mathbb Z_{(p_i)}$$ However the $\mathbb Z$-module $M$ is neither finitely generated nor projective (over $\mathbb Z$, projective=free)

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 $f_1,\cdots,f_n\in R$ with:  
$\alpha$)  $\sum Rf_i=R$    
$\beta$)   $M_{f_i}$ is a free $R_{f_i}$-modules of rank $1$ for all $i$.  

These modules are called **locally free of rank one**.   
**Remark:** $\beta$) implies that locally free modules of rank one  are finitely generated over $R$, since "finitely generated" is a local condition   

4) The locally free modules of rank one defined in 3) can also be characterized as the  modules $M$ over $R$ such that  equivalently:   
 i) The module $M$ is finitely generated, projective and for all primes $\mathfrak p\subset R$ the (necessarily!) free $R_\mathfrak p$- module $M_\mathfrak p$   has  rank $1$   
ii) The module $M$ is finitely generated and 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, but the conceptual definition is that given in 2) and 3).  

**Edit: WARNING**   
The confusion is made worse by Bourbaki's unfortunate decision to define  a projective module of rank $1$ as a **finitely generated** module $P$ for which $M_\mathfrak p$  is free of rank $1$ over $R_\mathfrak p$ for all primes $\mathfrak p\subset R$.  
As my example 1) $\beta$) shows, omitting to require that $P$ be finitely generated [as is done in Pete's condition (2) b)] means accepting modules which aren't even projective, and which  don't satisfy (2) a) nor (2) c) of the question.