Let $L$ be a sheaf of sets on some site $S$. Let $F$ be the presheaf obtained by composing $L$ with the free Rmodule functor, i.e. for any object $U$, we define $F(U)$ to be the free $R$module on the set $L(U)$. Is $F$ a sheaf?

Suppose $S$ is a topological space which has two nonempty open subsets $U$ and $V$ with an empty intersection. Let $L$ be the sheaf of sets that assigns to each open set the singleton set $\lbrace *\rbrace$. Let $R$ be the integers. Then the presheaf you describe assigns the integers to every open set, but this is not a sheaf because, for example, the section 2 (over $U$) and the section 3 (over $V$) agreee by default on $U\cap V$ but can't be patched together. 

