Let $\alpha_1,...,\alpha_n$ be a basis for $V$ and $\phi_1,...,\phi_n$ be a basis for $V^*$. A couple of linear algebra texts (Axler and Kunze/Hoffman) introduce the dual basis and say that the fact that "there is for each $i$ a unique linear functional $\phi_i$ on $V$ such that $\phi_i(\alpha_j) = \delta_{i,j}$" is implied by the theorem below (from Axler's "Linear Algebra").
I don't understand how the theorem in the image above implies the theorem in quotes. One reason that I don't understand how it follows is because the way that I would apply the theorem in the image is:
If $\alpha_1,...,\alpha_n$ is a basis for $V$ and $\phi_1,...,\phi_n$ is a basis for $V^*$, then exists a unique $T$ such that $T\alpha_i = \phi_i$.
However, this application of the theorem does not match the form of the quoted theorem. If instead of applying the theorem in the image to every basis vector of $V$ simultaneously, I instead apply it to each basis vector $\alpha_i$, then I would let $\alpha_i$, $F$, and $\phi_i$ take the place of $V$, $W$, and $T$ in the image, respectively. Then I would have that $\phi_i(\alpha_i) = 1$. Is this how the theorem in the image is being applied to imply the quoted theorem? If so, does the theorem in the image also imply that $\phi_i(\alpha_j) = 0$ when $i \neq j$? Or does that property only follow from the definition of the dual basis?
The Theorem 3.5 you are quoting is often used in the case where the $w_i$ form a basis, but that is not a hypothesis of the theorem and indeed the theorem is true for any choice of $w_i$.
In your case you can take $W = F$ (the field as a $1$-dimensional vector space) and $w_j = \delta_{ij}$. This means your linear functional is defined by $\phi_i(\alpha_i) = 1$ and $\phi_i(\alpha_j) = 0$ whenever $j \neq i$.
So in a sense, the condition $\phi_i(\alpha_j) = 0$ when $i \neq j$ does follow from the theorem and also follows from the definition of the dual basis. The theorem tells you that you can define a linear map by picking out the images of the basis elements. The definition of the dual basis is that for $\phi_i$ we choose $\delta_{ij}$ as the image of $\alpha_j$. The theorem says you can choose anything you want, the definition tells you the name we've given to one particular choice you could have made.