Map $f:M \to M^{\vee \vee}$ to Double Dual Space

Question

Let $R$ be a ring and $M$ a coherent $R$-module, therefore there exist a finite presentation: an exact sequence

$$R^I \to R^J \to M \to 0$$ (*)

with finite index sets $I,J$.

Let consider the canonical morphism $f:M \to M^{\vee \vee}$. My question is what can we say about injectivity and surjectivity of $f$?

I guess we can consider two cases:

1. $M$ has a "nice" classifacation theorem for finite generated modules (e.g. if $R$ is DVR or PID) such that we can assume $M= R^L \oplus T$ where $R^L$ is the free summand and $T$ the torsion summand.

Since every $\phi \in M^{\vee}= Hom_R(M,R)$ maps torison elements to zero we can conclude $M^{\vee}= R^L$ and therefore also $M^{\vee \vee}= R^L$. Therefore $f$ is always surjective. (Remark: This argument can only run if $M$ is finitely generated).

Is the proof ok? So is that indeed true that in case 1. $f$ is always surjective and injective iff $M$ is free?

2. $R$ is arbitrary so there is no theory allowing to split $M = F \oplus T$ into free and torsion summands. What I can say here about $f$?