Let $V$ a $\mathbb{F}$-space, and define the function $\Phi:V\rightarrow V^{**}$ by
$$\Phi(v)(f)=f(v)\quad\; \textrm{for all}\;\quad v\in V,\; f\in V^{*} $$
I didn't understand that definition, because it's seen that $\Phi$ is defined like it was $\Phi:V\times V^{*}\rightarrow V^{**}$. Do I have a different $\Phi$ for each $f$? (I think the answer is yes).
Now, let $\alpha=(v_{i})_{i\in I}$ a linearly independent family in $V$. Define the nullifier of $\alpha$ by
$$\alpha^{0}=\{f\in V^{**}:f(v_{i})=0\;\forall i\in I\}. $$
I want to prove that $\textrm{Span}(\alpha)\subset\Phi^{-1}\left((\alpha^{0})^{0}\right)$.
By definition, $(\alpha^{0})^{0}=\{g\in V^{**}:g(f_{i})=0\;\forall i\in I\}$, where $f_{i}(v_{j})=\delta_{i,j}= \begin{cases} 1,\;\textrm{if}\; i=j \\ 0, \;\textrm{if}\; i\neq j \end{cases}.$
What can I do?
The double dual space $V^{**}$ is the dual of $V^*$ then and element $\psi \in V^{**}$ is a linear map $\psi \colon V^* \longrightarrow \mathbb{F}$.
For an element $v\in V$ we can define $f\in V^* \mapsto f(v) \in \mathbb{F}$ which gives and element of $V^{**}$. Therefore we define the map $$ \Phi \colon V \longrightarrow V^{**} $$ by $$ \Phi(v) \colon V^* \longrightarrow \mathbb{F}, \, f \mapsto f(v) $$
For the second part, and element of $V^{**}$ may act on $V^{*}$ not on $V$ unless you give more structure.