Let $A$ be an associative algebra (or a ring) and $M$ a free finite-dimensional left $A$-module. Then there exists a right $A$-module structure with $M^\perp$ (the dual space of $M$) as the underlying space, equipped with the action defined by \begin{equation} (\varphi a)(m) := \varphi(a m) \end{equation}
for all $\varphi \in M^{\perp}$, $a \in A$ and $m \in M$. Since $(M^{\perp})^\perp \cong M$, I'm wondering if the original action $a \cdot m$ on $M$ can be re-obtained via the definition of the action on the dual space twice. I've attempted this myself by substituting $\varphi^{\perp} \in M$ for $\varphi$ and $m^{\perp} \in M^{\perp}$ for $m$ in the centred equation above, however this obtains $\varphi^{\perp}(a m^{\perp}) = a\varphi^{\perp}( m^{\perp}) = a \delta_{\varphi, m}$, which equals $a$ if and only if $\varphi = m$ and zero otherwise (assuming the bases of $M$ and $M^{\perp}$ are chosen appropriately); so perhaps I'm missing something.
I've tagged category theory, as this comes up in the dual equivalence of categories $\text{mod}(A)$ and $\text{mod}(A^{\text{op}})$ under the standard duality functor $D$.
The canonical bijection $F:M\to (M^\perp)^\perp$ is defined by $F(m)(\varphi)=\varphi(m)$ for $m\in M$ and $\varphi\in M^\perp$ (so, $F(m)$ is the homomorphism $M^\perp\to A$ that sends $\varphi$ to $\varphi(m)$). So now, your question is just whether $F$ is $A$-linear. Plugging in the definitions, $$F(am)(\varphi)=\varphi(am)=(\varphi a)(m)=F(m)(\varphi a)=(aF(m))(\varphi),$$ where at the end $aF(m)$ is defined using the $A$-module structure of the double dual $(M^\perp)^\perp$. Thus $F(am)=aF(m)$, so $F$ is indeed $A$-linear.