The "complete" Correspondence Theorem is imo both one of the most astonishing, beautiful theorems in the whole of mathematics (and in group theory, in particular), and also one of the most overseen/forgotten/unknown/"what?" ones. I think we could put it as follows:
Theorem: Let $\,G\, $ be a group, $\,N\lhd G\,\;$ . Then, there exists a $\,1-1\,$ correspondence between the subgroups of the quotient $\,G/N\,$ and the subgroups of $\,G\,$ that contain $\,N\,$ given by
$$\bar H\le G/N\mapsto H:=\{\;x\in G\;;\;xN\in \bar H\;\}\;,\;\;\text{and its inverse correspondence}$$
$$N\le H\le G\mapsto \bar H:=\{\;xN\in G/N\;;\;x\in H\;\}\;.\;\;\text{ and s.t.: }$$
$$\begin{align*}(1)&\bar H\lhd G/N\iff H\lhd G\\
(2)&[G/N:\bar H]=[G:H]\end{align*}$$
Thus, we can write $\,\bar H=H/N:=\{\;hN\;;\;h\in H\;\}$
In the above theorem we could talk of a general group epimorphism $\,\phi : G\to K\,\;,\;\;N:=\ker\phi$ ,
and then subgroups of $\,K\,$ are in $\,1-1\,$ correspondence with subgroups of $\,G\,$ that contain $\,N\,$ and this correspondence keeps normality and index.
The above addresses, I think, your doubts about (2) in your question.
BTW, the demonstration of the Correspondence Theorem is painfully simple...
$(5)$ If $\;[G:H] = 2,\,$ then by definition of the index of $H$ in $G$, there exists exactly two distinct left cosets: $H$ and $gH$, for $g \in G \setminus H.\;$ Now, for $g\in G\setminus H$, the right coset $Hg \neq H$. We know the left cosets of $H$ partition $G$, as do the right cosets of $G$. It follows, necessarily, that $gH=G\setminus H=Hg$, so any left coset is also a right coset in any subgroup of index $2$
For $(6)$: Yes, indeed. Spot on.
$(7)$ Yes, the usual notation for the factor group (also sometimes called a quotient group) is $G/N$. The integers modulo $n = \mathbb{Z}/n\mathbb{Z}\,$ which is simply a particular example of such a group, where the group $G = \mathbb Z$ and $N = \mathbb n\mathbb Z$.
Best Answer
First of all, if $H \trianglelefteq G$, and $x+H,y+H \in G/H$, then $x+H = y+H \Leftrightarrow x-y \in H$ (using additive notation).
Two elements $n+\mathbb{Z}, m+\mathbb{Z} \in \mathbb{Z} / 4\mathbb{Z}$ are equal iff $n-m \in 4\mathbb{Z}$, that is, $4 \mid n-m$ or equivalenty $n \equiv m\ \text{mod}\ 4$, thus, the elements of $\mathbb{Z} / 4\mathbb{Z}$ are the congruence classes $\text{mod}\ 4$. This also tells you the index.
What can you say about the structure of $\mathbb{Z} / 4\mathbb{Z}$? Hint: $1+1+1+1 \equiv 0 \ \text{mod}\ 4$.