Let $R$ be a commutative ring with unity. Let $R$-Mod denote the category of $R$-modules, and $Ab$ denote the category of Abelian groups. Now, it is known that a covariant additive functor
$T: R$-Mod $\to $ Ab preserves direct limits (resp. inverse limits) if and only if for some $B\in R$-Mod, we have $T(-)\cong B \otimes_R (-)$ (resp. $T(-)\cong Hom_R(B,-)$ ).
My question is the following : Let $T:R$-Mod $\to R$-Mod be a covariant linear functor. Is it true that $T$ preserves direct limits (resp. inverse limits) if and only if for some $B\in R$-Mod, we have $T(-)\cong B \otimes_R (-)$ (resp. $T(-)\cong Hom_R(B,-)$ ) ?
Yes, if $F:R\text{-mod}\to R\text{-mod}$ is an $R$-linear functor, then it preserves all colimits iff it is isomorphic to a functor of the form $B\otimes_R-$, and preserves all limits iff it is isomorphic to a functor of the form $\text{Hom}_R(B,-)$. The proofs are pretty much the same as for additive functors $R\text{-mod}\to\text{Ab}$. Briefly:
For the colimit-preserving case, let $B=FR$. There is a natural transformation $\alpha:FR\otimes_R-\to F$ given by the chain of natural maps $$\alpha_X:FR\otimes_RX\cong FR\otimes_R\text{Hom}_R(R,X)\stackrel{\text{id}\otimes F}{\longrightarrow}FR\otimes_R\text{Hom}_R(FR,FX) \stackrel{\text{eval}}{\longrightarrow}FX.$$
It is straightforward to check that $\alpha_X$ is an isomorphism for $X=R$. Since both $FR\otimes_R-$ and $F$ preserve coproducts, $\alpha_X$ is an isomorphism for $X$ any free module. Since any module $X$ fits in an exact sequence $P_1\to P_0\to X\to0$ with $P_1$ and $P_0$ free, and since both $FR\otimes_R-$ and $F$ are right exact, $\alpha_X$ is an isomorphism for all $X$, and so $\alpha$ is an isomorphism of functors.
For the limit-preserving case, this proof works:
The conditions of the Special Adjoint Functor Theorem are satisfied, and so $F$ has a left adjoint $L$. Then there are natural isomorphisms $$FX\cong\text{Hom}_R(R,FX)\cong\text{Hom}_R(LR,X),$$ and so $F\cong\text{Hom}_R(LR,-)$.