First, remember that a group is just a set together with a binary operation that satisfies certain conditions. Certainly, $G/N$ is a set; so the only question is whether the operation defined on it makes it a group.
The definition of the operation that we use on $G/N$ is the following:
if $gN$ and $hN$ are cosets, then we define the "product" of $gN$ and $hN$ to be the coset $ghN$.
So here, to add $\{0,3\}$ and $\{1,4\}$, we pick one element from each set, add them, and then look for the coset which contains the result. So we can take $0$ and $1$, and look for the coset that contains $0+1=1$, namely $\{1,4\}$. So the result of adding the coset $\{0,3\}$ with $\{1,4\}$ is $\{1,4\}$.
What if you picked a different element from each set? It doesn't matter: you get the same answer in the end:
- If you pick $0$ and $4$, you look for the coset of $0+4=4$, which is $\{1,4\}$;
- If you pick $3$ and $1$, you look for the coset of $1+3 = 4$, which is $\{1,4\}$;
- If you pick $3$ and $4$, you look for the coset of $3+4=1$, which is $\{1,4\}$.
Likewise, $\{0,3\} +\{2,5\} = \text{the coset of }0+2 = \{2,5\}$; and $\{1,4\}+\{2,5\}=\text{the coset of }1+2 = \{0,3\}$. You can check that this makes the set $G/N$ into a group.
But really, you don't want to think of the elements of $G/N$ as the sets. You want to think of them as "equivalence classes", and specifically, as "the equivalence class of an element of $G$".
So you don't want to think of $\{0,3\}$ as "the set whose elements are $0$ and $3$", you want to think of it as $[0]$, the equivalence class of $0$ (also of $3$). And you want to think of $\{1,4\}$ as $[1]$ or as $[4]$; and $\{2,5\}$ as $[2]$ or $[5]$. Then the addition is given by
$$[a]+[b] = [a+b].$$
The fact that the subgroup is normal is what guarantees that this is well defined: each element has two "names", but the name you pick does not affect the result you get.
Let me begin with a geometric example:
$\Bbb C^*$ be the group of non-zero complex numbers under multiplication
Let $H=\{x+iy\in \Bbb C:x^2+y^2=1\}$ which basically contains all the complex numbers lying on the unit circle.
Now consider the left coset $(3+4i)H$, Geometrically speaking this coset contains all points lying on the circle centered at the origin and radius $5$
(Why?)
In general the left coset $(a+ib)H$ contains all points on the circle centered at the origin and radius $\sqrt{a^2+b^2}$
Here the cosets of $H$ partition the punctured complex plane (i.e. $\Bbb C^*$) into concentric circles. Now isn't that amazing?
As seen in the above example:=
The essence of cosets lies in the fact that they partition the entire group into equivalence classes.
Moreover if the group is finite, each of the partitions will have the same number of elements, this is because of the way we define Right/Left cosets.
All the above observations leads to one of the most important result in Group Theory - The Lagrange's Theorem.
Best Answer
If you "multiply some element of $H$ on the left by some element from the group $G$", that is not a coset. If you multiply all elements of $H$ on the left by one element of $G$, the set of products is a coset.
If $H$ happens to be a normal subgroup (i.e. its left cosets are the same as its right cosets), then one can actually multiply cosets, and that gives another group, the quotient group $G/H$.
(I'm having trouble figuring out what you're trying to say in your last paragraph.)