Perhaps an example:
$\mathbb Z^+$ is the integers with the group operation of addition.
$6\mathbb Z$ is the integers that are multiple of 6. i.e. $\{\cdots, -12,-6,0,6,12,\cdots\}$
This is a subgroup of $\mathbb Z^+$
And since $\mathbb Z^+$ is an abelian group, $6\mathbb Z$ is a normal subgroup.
If we choose an element from $\mathbb Z$, say $1,$ and add it to every element in $6\mathbb Z$ we get the set $\{\cdots, -5, 1, 7, \cdots\}$ This is a coset of $6\mathbb Z.$ If we chose 7, we would get the same coset.
If we did this with an element of $6\mathbb Z$ we would get $6\mathbb Z.$
If we did this with other integers, we would get other cosets. However, there are only finitely many cosets.
$\{\cdots, 0, 6, \cdots\},\{\cdots, 1, 7, \cdots\},\{\cdots, 2, 8, \cdots\},\{\cdots, 3, \cdots\},\{\cdots, 4, \cdots\},\{\cdots, 5, \cdots\}$
As you note above, this is an equivalence relation.
We can choose one integer from each equivalence class, to represent the class.
This gives us a map (homomorphism) from the integers to the set $\{0,1,2,3,4,5\}.$
This is the factor group $\mathbb Z/6\mathbb Z.$
Generalizing, if we have a subgroup, we can find some set of cosets. If the group is non-abelian, the left cosets may not equal the right cosets. However, if the subgroup is normal, they will. The set of cosets is a quotient group.
Your proof of (1) isn’t wrong, but you’re making things much harder than necessary. There is no reason to index $E$ injectively. For each $\alpha\in A$ we have $\alpha\in E_\alpha\in E$, so $\alpha\in\bigcup E$, and since $\alpha$ was an arbitrary element of $A$, it follows that $A\subseteq\bigcup E$. On the other hand, $E_\alpha\subseteq A$ for each $\alpha\in A$, so $\bigcup E\subseteq A$, and therefore $\bigcup E=A$.
Now suppose that $E_\alpha,E_\beta\in E$ are such that $E_\alpha\cap E_\beta\ne\varnothing$. Let $a\in E_\alpha\cap E_\beta$; then $a\sim\alpha$ and $a\sim\beta$. Let $x\in E_\alpha$; then $x\sim\alpha\sim a\sim\beta$, so $x\sim\beta$, and therefore $x\in E_\beta$. This shows that $E_\alpha\subseteq E_\beta$, and a similar argument shows that $E_\beta\subseteq E_\alpha$ and hence that $E_\alpha=E_\beta$. Thus, $E$ is a partition of $A$.
The proof of (2) is better in that respect, but one needn’t work quite that hard to show that the $\sim$-equivalence classes are the parts of the original partition. Fix $a\in A$; $P$ is a partition of $A$, so there is a unique $S_a\in P$ such that $a\in S_a$. Now let $x\in A$: clearly $x\in E_a$ iff $x\sim a$ iff there is an $S\in P$ such that $\{x,a\}\subseteq S$ iff $x\in S_a$, since $S_a$ is the only member of $P$ containing $a$. But $x\in E_a$ iff $x\in S_a$ simply means that $E_a=S_a$. That is, for each $a\in A$ the $\sim$-class of $a$ is identical to the unique member of $P$ containing $a$, so the partition of $A$ into $\sim$-classes (which we know from (1) is a partition) is identical to the partition $P$.
Your understanding of the sense in which the notions of equivalence relation on $A$ and partition of $A$ are ‘the same’ is correct. The demonstration of equivalence isn’t actually complete, because (1) doesn’t go quite far enough: now that we have (2), we should show that if we start with an equivalence relation $\sim$ on $A$, produce the associated partition as in (1), and then use (2) to produce an associated equivalence relation, we get back the original equivalence relation $\sim$. However, this is trivial, since two equivalence relations that have the same equivalence classes are easily seen to be the same relation.
This old answer of mine is another discussion of these matters.
Best Answer
If $\sim$ is an equivalence relation, then we know it is a reflexive, symmetric, and transitive relation. Only then can say that $\sim$ partitions $E$ into equivalence classes. Let $x \sim y$. Then for any $z\in \bar x$, we know that $x\sim z$ and $y \sim x$ (by the symmetry), and then $y\sim z$ (transitivity). Therefore, if $x \sim y$ then $\bar x = \bar y$.