Let $E$ be a vector space and $(x_i^*)_{1\leq i\leq p}$ a family of elements of $E^*$. Write $G'$ for the subspace generated by the $x^*_i$ and let $F$ be the orthogonal of $G'$ in $E$. Assume the $x^*_i$ are linearly independent. We want to show the existence of a family $(x_i)_{1\leq i\leq p}$ of elements of $E$ such that $\langle x_i,x^*_j\rangle=\delta_{ij}$ for all $i,j$. There author says that
...$E/F$ and $G'$ can each be canonically identified with the dual of the other; if the family $(x^*_i)$ is free, there is in $E/F$ a basis $(\dot{x}_i)$ the dual of $(x^*_i)$ and every representative system of the classes $\dot{x}_i$ has the required properties.
Let $\pi:E\rightarrow E/F$ be the canonical linear surjection. Then $^t\pi:(E/F)^*\rightarrow G'$ is an isomorphism (the transpose of $\pi$). This shows that we can identity $(E/F)^*$ and $G'$. But how should we identify $G'^*$ and $E/F$? Taking the transpose again gives another isomorphism $^{tt}\pi:(G')^*\rightarrow(E/F)^{**}$. This means that $(E/F)^{**}$ is finite dimensional and so the canonical mapping $c_{E/F}:E/F\rightarrow (E/F)^{**}$ is an isomorphism. Does the author mean that $$\dot{x}_i:=c_{E/F}^{-1}(^{tt}\pi(x^{**}_i)),$$ where $(x^{**}_i)$ is the dual basis of $(x^{*}_i)$?
The inclusion $i : G' \hookrightarrow E^*$ yields $^ti : E^{**} \to (G')^*$, and then $^ti \circ c_E : E \to (G')^*$ where $c_E : E \to E^{**}$ is the canonical injection. Now note that \begin{align} \ker({}^ti \circ c_E) &= \{x \in E : c_E(x) \circ i = 0\} \\ &= \{x \in E : (\forall z \in G') \ c_E(x)(z)=0\} = F, \end{align} so that $^ti \circ c_E$ induces an isomorphism $E/F \to (G')^*$ that sends the class of $x \in E$ to $c_E(x) \circ i$, the restriction of $c_E(x)$ on $G'$.