$\newcommand{\Hom}{\operatorname{Hom}}$Let $U,V$ be vector spaces. Denote the dual space of $U$ with $U^\intercal$ and the dual mapping of a linear map $\Phi$ be $\Phi^\intercal$. Define the double dual embedding on $U$ to be $\iota_U:U \to U^{\intercal\intercal}$. ($u \in U, f \in U^\intercal$) $$ \iota_U(u)(f) = f(u) $$
If $\Phi \in \Hom(V,U)$, then $\Phi^{\intercal\intercal} \circ \iota_V = \iota_U \circ \Phi:V \to U^{\intercal\intercal}$
Proof: Let $v \in V, f \in U^\intercal$. Then $$ \iota_U \circ \Phi(v)(f) = f(\Phi(v)) $$ but how do I show that $$ \Phi^{\intercal\intercal} \circ \iota_V(v)(f) = f(\Phi(v)) $$ ?
What you need to show is that $\Phi^{TT}( \iota_V(v))(f)=f(\Phi(v))$. Notice that $\Phi^{TT}(\iota_V(v))=\iota_V(v)\circ \Phi^T$, which maps $f$ to $\iota_V(v)(\Phi^T(f))=\iota_V(v)(f\circ \Phi)=f(\Phi(v))$.