The issue with working with quotient groups is that there are many representatives of the same coset. For example, in $\mathbb{Z}/5\mathbb{Z}$ one has that
$$1+5\mathbb{Z}=\{\ldots,-9,-4,1,6,11,\ldots\}=6+5\mathbb{Z}$$
and
$$2+5\mathbb{Z}=\{\ldots,-8,-3,2,7,12,\ldots\}=12+5\mathbb{Z}.$$
It is, of course, reasonable to be concerned whether
$$3+5\mathbb{Z}=(1+5\mathbb{Z})+(2+5\mathbb{Z})=(6+5\mathbb{Z})+(12+5\mathbb{Z})=18+5\mathbb{Z}?$$
Of course, in this case everything works out just fine, but it is not always so. For example, take the subgroup $H=\langle(12)\rangle=\{(1),(12)\}\leq S_3$. We have
$$(13)H=\{(13),(123)\}=(123)H$$
and
$$(23)H=\{(23),(321)\}=(321)H$$
However, $(13)(23)H=(321)H$, while $(123)(321)H=(1)H=H$. Hence, $$(13)H(23)H=(13)(23)H=(321)H\neq H=(123)(321)H=(123)H(321)H$$ and the operation is not well defined.
The difference in the two cases is that $5\mathbb{Z}$ is a normal subgroup of $\mathbb{Z}$, while $H$ is not normal in $S_3$.
If we assume $H$ is a normal subgroup of $G$, we can show that the operation $aHbH=abH$ is well defined as follows:
Suppose $aH=cH$ and $bH=dH$. By definition, this means that $c^{-1}a\in H$ and $d^{-1}b\in H$. To show that $abH=cdH$, we need to show that $(cd)^{-1}(ab)\in H$.
Well,
$$
(cd)^{-1}ab=d^{-1}c^{-1}ab=(d^{-1}(c^{-1}a)d)(d^{-1}b).
$$
By assumption $d^{-1}b\in H$. Also, since $c^{-1}a\in H$ and $H$ is normal $d^{-1}(c^{-1}a)d\in H$. Finally, $H$ is a subgroup, so $(d^{-1}(c^{-1}a)d)(d^{-1}b)\in H$ and we're done.
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
The Wikipedia is using the (very) common convention of using multiplicative notation for a group, regardless of how the operation of a specific group may be defined or denoted. It would be more explicit to say, if $(G, *)$ is a group, the left cosets are defined by $g*H = \{g*h : h \in H\}$, etc.
So the operation used between an element $g$ and subgroup $H$ is whatever operation the group has, whether it's called multiplication, addition, composition, or some fourth thing. If you had a group with which you used additive notation, you'd write $g + H$ for cosets.
In theory you could do a similar thing, writing cosets as $g \circ H$ for groups whose operation is composition, but I can't say I've ever seen it. Often we just use multiplicative notation ("juxtaposition") even for groups whose operation is function composition. Really, you'll generally see juxtaposition used for almost every group that isn't written additively.