Can we construct the field extension not in the meaning of embedding? Here is my question. The book uses a method making $F$ embedded in the $K$, and thus regarding $F$ as a subfield of $K$. Instead of doing in this way, how can we really construct a field which contains $F$, not in the meaning of embedding?

 A: In general, requiring that things actually be subsets as opposed to being isomorphic to a subset is very restrictive. 
For example, is it true that $\mathbb{Z}\subseteq \mathbb{Q}$? Obviously we would like to say yes. However formally, elements of $\mathbb{Q}$ are actually equivalence classes of $(p,q)\in \mathbb{Z}\times (\mathbb{Z}-\{0\})$. So the $3\in \mathbb{Z}$ is not the same as the $3\in \mathbb{Q}$. Technically, what we refer to as "$3$" in $\mathbb{Q}$ is the set $\left[\frac{3}{1}\right]:=\{(3,1),(-3,-1),(6,2),(-6,-2),(9,3),\ldots\}$. But there is an embedding $\varphi:\mathbb{Z}\hookrightarrow \mathbb{Q}$ given by $\varphi(n)=\left[\frac{n}{1}\right]$. Not only is $\varphi$ injective, but it also preserves all the properties about $\mathbb{Z}$ that we care about (addition, multiplication, ordering, etc.). Therefore, we don't bother distinguishing between $\mathbb{Z}$ and $\varphi(\mathbb{Z})\subseteq \mathbb{Q}$. 
If you really wanted to, you could take extra care to always write $\varphi(3)\cdot \varphi(5)=\varphi(15)$ as opposed to $3\cdot 5=15$ when working in $\mathbb{Q}$. But there really wouldn't be much point; None of the math changes. The only thing that would be different is that your hand would get tired from writing all of the extra notation. In general, mathematicians don't care so much about the explicit objects but rather the properties that they have. Whether I think of $\mathbb{Z}_2$ as the set of permutations of the set $\{a,b\}$ or the set of permutations of the set $\{x,y\}$ or the set of equivalence classes $\{2\mathbb{Z},2\mathbb{Z}+1\}$, it doesn't matter. It's the same field in different clothes.
As for your question, we are merely identifying $F$ with the set of constant functions in $F[x]/\langle p(x)\rangle$. If you really don't think $F$ is the same as the constant functions, then we could take the union $F\bigcup\{f(x)+\langle p(x)\rangle \in F[x]/\langle p(x) \rangle:\mathrm{deg}(f)>0\}$ which would "actually" contain $F$. Then you would have to define the addition and multiplication all over again on this new set to show its a field. But that's a lot of unnecessary work which, in the end, would give us all the same answers as if we had taken the easy route. So my advice: take the easy route.
A: There's nothing special about fields here. The key point is that whenever I have a structure $\mathcal{A}$ with underlying set $A$ and a bijection $b$ between $A$ and some set $B$, I get an induced structure $\mathcal{B}$ with underlying set $B$ gotten by "pushing $\mathcal{A}$ along $b$." For example, suppose my language has a binary operation symbol "$*$" (like addition or multiplication in the field context); then given an interpretation of this symbol $*^\mathcal{A}$ on the $A$-side, we get a corresponding operation on the $B$-side given by $$x*^\mathcal{B}y=b(b^{-1}(x)*^\mathcal{A}b^{-1}(y)).$$

This can be used to "massage embeddings into inclusions." Specifically, suppose we have a structure $\mathcal{E}$, another structure $\mathcal{F}$, and an embedding $i:\mathcal{E}\rightarrow\mathcal{F}$. Let $E$ and $F$ be the underlying sets of $\mathcal{E}$ and $\mathcal{F}$ respectively; without loss of generality, assume $E\cap F=\emptyset$ (otherwise "disjointify" appropriately). Now consider the set $$D=E\cup(F\setminus i[E]).$$ 
There is a natural bijection $b:F\rightarrow D$ given by


*

*$b(x)=x$ if $x\in F\setminus i[E]$, and

*$b(x)=i^{-1}(x)$ if $x\in ran(i)$.
It's not hard to show that the structure $\mathcal{D}$ on $D$ induced by $b$ and $\mathcal{F}$ has the desired properties, specifically:


*

*$b$ is an isomorphism between $\mathcal{F}$ and $\mathcal{D}$.

*$\mathcal{E}\subseteq\mathcal{D}$, and $b\circ i$ is the inclusion map (so $\mathcal{E}$ is a substructure of $\mathcal{D}$ "in the same way" that $\mathcal{E}$ was originally embedded into $\mathcal{F}$).
