See my answer to the question about construction of opposite categories. There I descripe the dual category of $R \text{-Mod}$ geometrically. But to be honest, I don't think that this description is useful at all. Note that $(R \text{-Mod})^{op}$ is never equivalent to some $S \text{-Mod}$, since in $R \text{-Mod}$ we have $\cup_i (F_i \cap G) = (\cup_i F_i) \cap G$ for a filtered system $F_i$, but we do not have the dual statement $\cap_i (F_i + G) = (\cap F_i) + G$.
Here is another example: Let $R$ be a PID, which is not a field. Then $Hom(-,Q(R)/R)$ yields an equivalence between the abelian category of finitely generated $R$-modules and its dual. Perhaps this also works for hausdorff locally compact abelian groups (Pontryagin duality); but perhaps this category is not abelian: Kernels and Cokernels exist by abstract nonsense, but I don't think that every epimorphism is a cokernel.
See my answer to the question about construction of opposite categories. There I descripe the dual category of $R \text{-Mod}$ geometrically. But to be honest, I don't think that this description is useful at all. Note that $(R \text{-Mod})^{op}$ is never equivalent to some $S \text{-Mod}$, since in $R \text{-Mod}$ we have $\cup_i (F_i \cap G) = (\cup_i F_i) \cap G$ for a filtered system $F_i$, but we do not have the dual statement $\cap_i (F_i + G) = (\cap F_i) + G$.
Here is another example: Let $R$ be a PID, which is not a field. Then $Hom(-,Q(R)/R)$ yields an equivalence between the abelian category of finitely generated $R$-modules and its dual.