Normal extensions equivalence, Morandi Galois Theory, p36, Theorem 3.28 
If $K$ is algebraic over $F$, then the following are equivalent: 
(2) If $M$ is an algebraic closure of $K$ and if $\tau: K \rightarrow M$ is an $F$-homomorphism, then $\tau(K)=K$. 
(3) If $F \subseteq L \subseteq K \subseteq N$ are fields and if $\sigma: L \rightarrow N$ is an $F$-homomorphism, then $\sigma(L) \subseteq K$, and there is a $\tau \in Gal(K/F)$ with $\tau_L = \sigma$. 

--- I have a number of questions regarding the proof (highlighted)---

Proof: Suppose $F \subseteq L \subseteq K \subseteq N$, $\sigma:L \rightarrow N$ is an $F$-homomorphism. Since $L \subseteq K$, the extension $L/F$, the extension $L/F$ is algebraic, and $\sigma(L) \subseteq N$ is algebraic over $F$. Let $M'$ be the algebraic closure of $F$ in $N$ (1) and let $M$ be the algebraic closure of $M'$. By the isomorphism extension theorem(2) there is an extension $\rho:M \rightarrow M$ with $\rho|_L= \sigma$. Let $\tau= \rho|_K$. By 2 we have $\tau(K)=K$, so $\sigma(L) = \tau(L) \subseteq \tau(K)= K$. Thus $\tau \in Gal(K/F)$. 


(1) What does this mean? This was not referred anywhere in the text. Do we define $M':=\{ \alpha \in N \,:\, \alpha \text{ is algebraic over  } M \}$? 
(2) I do not see how the theorem is applied. The theorem is stated as 

 A: (1) The algebraic closure $M'$ of $F$ in $N$ is $$M':=\{\alpha\in N\mid \alpha\text{ is algebraic over} F\}=M\cap N.$$ Note that $N$ may not be algebraic over $F$, and the $F$-embedding $\sigma$ is actually $\sigma\colon L\to M'\subset M$, into an algebraically closed field containing $L$.
(2) I think the theorem you cited is not sufficient for the deduction (if it is stated for a finite family of polynomials), and it would be much clearer for this deduction if you use the following

Theorem. Let $L/F$ be an algebraic extension and $M$ the algebraic closure of $L$. If $\sigma\colon L\to M$ is an $F$-embedding, and $E/L$ is algebraic, then $\sigma$ can be extended to an $F$-embedding $\tau\colon E\to M$.

Proof of Theorem. Consider the set $S$ consisting of pairs $(k,\varphi)$ where $k$ is an intermediate field $L\subseteq k\subseteq E$ and $\varphi\colon k\to E$ is an $F$-embedding such that $\varphi|_L=\sigma$. Define $(k_1,\varphi_1)\geq (k_2,\varphi_2)$ iff. $k_1\supseteq k_2$ and $\varphi_1|_{k_2}=\varphi_2$. Then $(S,\geq)$ becomes a poset and each chain (totally ordered subset) $\{(k_\lambda,\varphi_\lambda)\}_{\lambda\in\Lambda}$ of $S$ has an upper bound $(\cup_{\lambda\in\Lambda} k_\lambda, \cup_{\lambda\in\Lambda} \varphi_\lambda)$ where $\cup_{\lambda\in\Lambda} \varphi_\lambda$ is defined by $\cup_{\lambda\in\Lambda} \varphi_\lambda|_{k_{\lambda'}}=\varphi_{\lambda'}$. Invoking Zorn's lemma one has a maximal element $(k',\tau)$ in $S$. Then $k'=E$, which can be seen from contraposition: if $k'$ is properly contained in $E$, then one can find an $\alpha\in E\setminus k'$, and the Theorem 3.20 in your post provides a pair $(k'',\varphi'')>(k',\tau)$  (while in fact one only needs to adjoin an element $\alpha\in E\setminus k'$ to $k'$ and extend $\tau$ to some $F$-embedding $\tau'\colon k'(\alpha)\to M$ to get the pair $(k'',\varphi'')$). Q.E.D.
By the Theorem above you can immediately figure out how the questioned deduction in your post is made. 
ps. Besides Lang's Algebra, as DonAntonio recommended, Jacobson's Basic Algebra I is also a very good reference for you to learn Galois theory.
