For a representation $r\colon G\to GL(V)$, the dual representation is $\phi^*$ defined as $g \mapsto \rho^*(g^{-1})$ where $\rho^*(g) (f) = f\circ \rho(g)$ i.e. for all $v\in V$, $\rho^*(g) (f) (v) = f([\rho(g)](v))$.
So $\rho^*(g) \colon f_1 \mapsto f_1\circ \rho(g^{-1})$ for $f_1\colon V\to \mathbb C$.
So far this seems understandable. But can one express the double dual representation $\rho^{* *}(g)$ using $\rho(g)$ (for all $g\in G$)? What about when $f_2$ is a linear functional on $V^{*}$, $\rho^{**}(g) \colon f_2\mapsto f_2\circ \rho^{*}(g^{-1}) =?$
In other words, given a linear functional $f\colon V^{*} \to \mathbb C$, what is the action of $g$ on this element $f$ in $V^{**}$ (in terms of $\rho(g)$) ?
I am assuming you are talking about finite dimensional representations.
There is a canonical isomorphism of vector spaces $\Phi\colon V\xrightarrow{\cong}V^{**}$ given by $\Phi(v)(f)=f(v)$, i.e., $\Phi(v)\in(V^*)^*$ sends $f\in V^*$ to its value on $v$. This is standard and can be found in linear algebra books.
I will denote the $G$-actions by just writing $g(-)$. So the action of $g$ on $v\in V$ is given by $gv$, that on $f\in V^*$ by $gf=f(g^{-1}(-))$.
For $g\in G$ we have $$ (g\Phi(v))(f)=\Phi(v)(g^{-1}f)=f((g^{-1})^{-1}v)=f(gv)=\Phi(gv)(f), $$ hence $g\Phi(v)=\Phi(gv)$. This shows that $\Phi$ is $G$-equivariant.
I hope that helps.
Note: This sort of result applies more generally to modules over Hopf algebras, at least if they are cocommutative or have other suitable structure