I don't understand the proof of the second isomorphism theorem. 
I don't understand how the author gets the last couple of lines. I understand, by the homomorphism theorem(perhaps sometimes called "The First Ismoprhism Theorem"), $\bar{G} \cong G/N$, but how do they then conclude $(G/N)/(K/N) \cong  G/K$?
Is the idea to show  $\bar{K} \cong (K/N)$ first!? If so, how? If this is the case, I would greatly the details. Also, if this is indeed the case, wouldn't they also have to show $A \cong B$ and $C \cong D \implies A/C \cong B/D?$
If the idea is not to show  $\bar{K} \cong (K/N)$, please let me know.
 A: Let $f \colon \overline{G} \to G/N$ denote the isomorphism guaranteed by the First Isomorphism Theorem. When the author writes "this identification carries $\overline{K}$ to the image of $K$ in $G/N$, namely $K/N$" what they mean is that
$$f(\overline{K}) = K/N \subset G/N.$$
You should check that this is true, and that in fact the restriction $f|_{\overline{K}} \colon \overline{K} \to K/N$ is a group isomorphism as well.
Hint: You should write down explicitly what $f$ does, and be careful checking that various things are well-defined.

In general it is not true that $A \cong B$ and $C \cong D$ imply $A/C \cong B/D.$ However, in this case, since one isomorphism $f \colon \overline{K} \to K/N$ is the restriction of the other $f \colon \overline{G} \to G/N$, this statement is true. Let's prove this:
Lemma If $A, C$ are groups and $B \subseteq A$, $D \subseteq C$ normal subgroups, and if $f \colon A \to C$ is an isomorphism such that $f|_{B} \colon B \to D$ is also an isomorphism, then
$$A/B \cong C/D.$$
Proof: The obvious isomorphism $\overline{f}$ is as follows: you take an element $aB$ of $A/B$, and you send it to $f(a)D.$ But we need to check that this is well-defined.
If you give me $a, a' \in A$ such that $aB = a'B$, is it true that $f(a)D = f(a')D?$ Well, if $aB = a'B$, then $a^{-1}a' \in B,$ and thus $f(a^{-1}a') \in f(B) = D$. Therefore, we have
$$f(a')D = f(a)f(a)^{-1}f(a')D = f(a)f(a^{-1}a')D = f(a)D,$$ where $f(a^{-1}a')D$ is the identity because $f(a^{-1}a') \in D.$
So, $\overline{f}$ is indeed well-defined. Why is it an isomorphism?
It is a group homomorphism because given any $a, a' \in A$, we have
$$\overline{f}(aBa'B) = \overline{f}((aa')B) = f(aa')D = f(a)f(a')D = f(a)Df(a')D = \overline{f}(aB)\overline{f}(a'B).$$
It's injective because if $aB \in A/B$ is sent to the identity $D$ of $C/D$, then we have
$$D = \overline{f}(aB) = f(a)D,$$ hence $f(a) \in D$. Since $f$ restricts to an isomorphism from $B$ to $D$, it is a bijection from $B$ to $D$ as well, so it follows that $a$ must be in $B.$ So, we have $aB = B,$ which means that the only element sent to the identity $D$ is the identity $B$ of $A/B$.
It's surjective because for any $cD$ in $C/D$, there exists $a \in A$ such that $f(a) = c$ (because $f$ is an isomorphism), and thus $\overline{f}(aB) = cD.$
So, we've shown that $\overline{f}$ is an isomorphism, and we're done.

Now, we know that $f \colon \overline{G} \to G/N$ is an isomorphism, and it restricts to an isomorphism $\overline{K} \to K/N$. By the Lemma, we have
$$\overline{G}/\overline{K} \cong (G/N)/(K/N),$$ as desired.
A: First part of this proof is actually what you mentioned that $\overline{K} \cong (K/N)$. from first isomorphism theorem $\overline{\phi} : G/N \rightarrow \overline{G}$ is isomorphism (identification) and $\overline{\phi}^{-1} :\overline{G} \rightarrow G/N$ is the inverse homeomorphism (isomorphism). the proof shows that $\overline{\phi}^{-1}(\overline{K})=K/N.$ (actually shows the inverse but $ \overline{\phi}$ is isomorphism). 
last lines of the proof uses the fact that if $h:H\rightarrow H'$ is a isomorphism and $K \triangleleft H$ then $H/K \cong H'/h(K)$. author first show that $G/N \cong \overline{G}$. and then shows that this isomorphism maps normal subgroup $\overline{K}$ of $\overline{G}$ to normal subgroup $K/N$ of $G/N$. so the proof is completed. 
but if you're not comfortable with this proof you can do it in the hard way and use first isomorphism theorem directly.

In order to proof the theorem using first isomorphism theorem we should introduce a homeomorphism $q:\frac{G}{N} \rightarrow \frac{\overline{G}}{\overline{K}}$. we can do this using $\phi$ and $\psi$ as introduced in proof:
$$
\frac{G}{N} \xrightarrow{\overline{\phi}} \overline{G} \xrightarrow{\psi} \frac{\overline{G}}{\overline{K}} \\
xN \mapsto \phi(x) \mapsto \phi(x) \overline{K} 
$$
so $q=\psi \circ \overline{\phi} :\frac{G}{N} \rightarrow \frac{\overline{G}}{\overline{K}}$. from first isomorphism theorem we get :
$$
\frac{(G/N)}{ker(q)} \cong \frac{\overline{G}}{\overline{K}} \tag{1}
$$
we should determine $ker(q)$. 
$$
ker(q) = \{xN \in G/N \,|\, q(xN)= \phi(x) \overline{K}=\overline{K}\} = \{xN\in G/N \, | \, \phi(x) \in \overline{K} \} \\
=\{xN \in G/N \, | \, x \in K\} = K/N \tag{2}
$$
final thing you need is $\overline{G}/\overline{K} \cong G/K$. for this isomorphism you can use $\overline{\phi}$ again. (this is the part that author use $\psi \circ \phi$)
$$
\overline{\phi}': G/K \xrightarrow{\cong} \overline{G}/\overline{K} \\
xK\mapsto \overline{\phi}(x) \overline{K} \tag{3}
$$
put $(2)$ and result of $(3)$ in $(1)$ and you get :
$$
\frac{(G/N)}{(K/N)} \cong \frac{\overline{G}}{\overline{K}} \cong \frac{G}{K}
$$
