Lemma 5.14. says
Let $K(\alpha) : K$ be a simple algebraic extension, let the minimal polynomial of $\alpha$ over $K$ be $m$, and let $\partial m = n$. Then $\{1,\alpha,\ldots,\alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$.
Here $K$ is a field and $\{1,\alpha,\ldots,\alpha^{n-1}\}$ is a basis for $K(\alpha)$ over $K$ as a vector space. $\partial m$ is the degree of $m$. The proof given for lemma 5.14. is just
The theorem is a restatement of Lemma 5.9.
Lemma 5.9. says
Every polynomial $a \in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \partial m$.
Here $K[t]$ is a polynomial ring over the field $K$.
I understand the statements in both lemmas, but I can't figure out their connections. Lemma 5.9. is just about properties of polynomials and lemma 5.14. is about properties of simple algebraic extensions and their minimal polynomials.
Let us show Lemma 5.14, just see to what it amounts. We consider the system $S$ with the elements $1,\alpha,\dots,\alpha^{n-1}$, and have to show two properties of this system, seen as a subset in the (underlying) vector space of the given field $K(\alpha)$ over the field $K$.
the system $S$ is generating the vector space. For this let us pick some element in $K(\alpha)$. By its definition it is a finite linear combination of powers of $\alpha$, possibly some of them having powers bigger then $n$. At any rate, this element is of the shape $p(\alpha)$ for an appropriate polynomial $p\in K[X]$. By the Lemma 5.9 we know that we can write $p\cong r$ modulo $m$ with the rest $r$ having degree $<n$. Just as a remark, we use the existence part from Lemma 5.9, where "is congruent to" may be better replaced by "there exists a polynomial $r$ (congruent with the given one modulo $m$)"... Then we have a relation of the shape $p=q\cdot m+r$, so $$p(\alpha)=q(\alpha)\cdot m(\alpha)+r(\alpha)=q(\alpha)\cdot0+r(\alpha)=r(\alpha)\ .$$ Writing $r(\alpha)$ explicitly, we see that it is a $K$-linear combination of the elements in $S$. So $S$ is a generating system.
the system $S$ is linearly independent. Let us assume that we have a $K$-linear combination of the elements of $S$ equal to zero. We rewrite this as $r(\alpha)=0$, $r\in K[X]$ being of degree $<n$. Since $m$ is a monic polynomial of minimal degree with $m(\alpha)=0$, we obtain $r=0$, i.e. $r$ is the zero polynomial. (Else it has a principal coefficient $\ne 0$, and we can renorm $r$ by dividing by it, getting a monic polynomial of strictly smaller degree which annihilates $\alpha$. Contradiction.) Alternatively, we can put this in an other shape to look like Lemma 5.9. Let us write $r(\alpha)=0=0(\alpha)$. Here, the $0$ in $0(\alpha)$ is the zero polynomial, which also has degree $<n$. We use now the uniqueness in the Lemma 5.9, getting $r=0$ in $K[X]$.
So the essence of Lemma 5.14 means that