Let's look at a slightly more "in-depth" example: Consider the subgroup $7\Bbb Z$ of $\Bbb Z$ which consists of all (positive, negative and $0$) multiples of $7$. Let's "expand" this a bit:
$7\Bbb Z = \{\dots,-21,-14,-7,0,7,14,21,28,\dots\}$
What set do we get if we add $1$ to everything in this set?
$1 + 7\Bbb Z = \{\dots,-20,-13,-6,1,8,15,22,29,\dots\}$
If we shift "over $2$" we get:
$2 + 7\Bbb Z = \{\dots,-20,-12,-5,2,9,16,23,30,\dots\}$
Now here is a curious thing: If we take something in, say $3 + 7\Bbb Z$, and add it to something in $2 + 7\Bbb Z$, we'll get something in $5 + 7\Bbb Z $. For example:
$10 + 9 = 19$ and $19$ is $5$ more than $14$.
This process "chops up" $\Bbb Z$ into $7$ pieces (each of which is "infinite") and in every piece, the numbers are all "$7$ apart".
So the "$k$" in $k + 7\Bbb Z$ measures how far to the right (right = counting up) we are from a multiple of $7$.
If you imagine the integers as an infinite string of beads, we are "wrapping the beads into a circle" so that the $7$th bead winds up in the same place as the "$0$-bead". In fact, any two beads that are a multiple of $7$ apart on our original string, wind up in the same place on our "$7$ position circle".
You've seen this before: on a clock (there it's "modulo $12$" instead of $7$, or "modulo $24$" if you use military time).
This has the effect of effectively "setting all multiples of $7$ equal to $0$". Since we know that $14$ (for example) isn't actually $0$, we don't say:
$14 = 0$, but rather, $14$ is equivalent to $0$ since it is a multiple of $7$ away from $0$ (if $6$ was "as high as we could count before starting over", $14$ would actually BE $\ 0$).
It's somewhat "magical" that all this actually WORKS, in that we don't get any contradictions or confusion this way. Part of this has to do with the fact that addition is commutative: that is, that $k + m = m+k$ for any two integers $k,m$. With a group in general, we need a more special condition on the subgroup to make this all work out (it has to be normal).
It turns out that with a general group $G$ and a subgroup $H$, that the set:
$(aH)(bH) = \{(ah)(bh'): h,h' \in H\}$ doesn't always equal some other coset $kH$, so subgroups for which this DOES happen are "special".
If $G$ is abelian, of course, then $(ah)(bh') = a(hb)h' = a(bh)h' = (ab)(hh')$ and this DOES happen. So quotient groups of abelian groups are "easier to understand".
In this special case of $G = \Bbb Z$ and $H = ${even integers}, this process is called "the arithmetic of parity" (reasoning by evens and odds). This "trick" (reducing mod $2$) turns an "infinite number of cases" to just TWO cases, and often that is all we need.
Best Answer
First of all you have to ask yourself if it is possibile that $H/K$ is a group, i.e if $K$ is normal in $H$.
In your case (please verify this fact) you have that $K$ is an abelian group (in particular is isomorphic to $(\mathbb{R},+,0)$ ) and is contained in the center of $H$ (I.e $AK=KA$ for each $A \in H$, and $K\in K$), so that $K$ is clearly normal in $H$.
Thus The quotient set $H/K=\{AK: A \in H\}$ inherits also a group structure.
Now we have to understand what is the multiplication in $H$ to compute easily the quotient set $H/K$.
Taking an element $A\in H$, I will denote $A$ as $A(a,b,c)$, where $A_{12}=a, A_{13}=b$ and $A_{23}=c$.
Then taking two matrices $A=A(a,b,c)$ and $B=B(a’,b’,c’)$, you have
$AB=AB(a+a’, b+ac’ +b’, c+c’)=BA$
and so if you have a general element $A=A(a,b,c)$, you can observe that $A’(a,0,c) $ is an element in its class on $H/K$, in fact
$A(a,b,c)=A’(a, 0, c)K(0, b+ac’+b’, 0) $
So that you can define the section map $s: H/K \to G$ Sending each class $AK$ to $A’(a,0,c)$, where $(G:=\{ A(a,0,c) : a,c \in \mathbb{R}\},+,0)$ is an additive group.
You can prove that this map is an isomorphism of groups and observing that $A(a,0,c)+B(a’,0,c’)=(A+B)(a+a’, 0, c+c’)$
then you can say that $H/K \cong (\mathbb{R}^2,+,0)$