Let $G$ be a group, and $H\triangleleft G$ be a normal subgroup of $G$.
To my best understanding,the group $^G/_H$ is often used to "simplify" groups, meaning it enables us to deal with a simple form of $G$, so in other words, throwing all the elements in $H$ together and labeling them as $e_{G/H}$ (the unity of $^G/_H$), is to say that $H$ is, in some sense, "bothering us", or "interfering" when trying to investigate $G$.
In what way is $H$ "bad"?
I feel like there's something deep that I'm missing here...
Groups arise very naturally as symmetries of geometric objects, for example as in the comment of @MarcinLos.
More generally, groups arise as actions: a group $G$ can "act" on some geometric object. The difference between "symmetry groups" and "actions" is that when a group $G$ acts on a geometric object $X$, some group elements might act trivially on $X$, meaning that they act on $X$ as the identity map. The subset of $G$ which acts as the identity map on $X$ is a normal subgroup called the "kernel" of the action. The quotient group $G/$kernel can then be thought of as a true symmetry group, in which each group element acts nontrivially. So in this case the "bad" elements of $G$ are the elements of the kernel, the elements that are acting trivially on $X$.
There are zillions of examples of this phenomenon. Here is just one example, one of my favorites: the group $SL_2(\mathbb{R})$, which is the group of $2 \times 2$ matrices with real coefficients and determinant $1$. This group acts on $\mathbb{R}^* = \mathbb{R} \cup \{\infty\}$ by "fractional linear transformations": $$\begin{pmatrix}a&b\\c&d\end{pmatrix} r = \frac{ar+b}{cr+d} $$ (and if you ever learn about hyperbolic geometry $\mathbb{H}$ modelled as the upper half of the complex plane, one can turn this into an action on $\mathbb{H}$ by replacing the $\mathbb{R}^*$ variable $r$ with a complex variable $x+iy$, $y >0$).
Of course the identity matrix $I$ acts as the identity map on $\mathbb{R}^*$. More "bothersome" is that the negative of the identity matrix $$-I = \begin{pmatrix}-1&0\\0&-1\end{pmatrix} $$ also acts as the identity map on $\mathbb{R}^*$. Fortunately, without too much trouble one can prove that $\pm I$ is the kernel of the action: these are the only matrices which act as the identity map on $\mathbb{R}^*$.
So if you are bothered by $\{\pm I\}$ acting as the identity, you may want to "simplify" the action by taking the quotient group: $$PSL(2,\mathbb{R}) = SL(2,\mathbb{R}) / \{\pm I\} $$ And now you get a true symmetry group, each non-identity element of which acts on $\mathbb{R}^*$ (and on $\mathbb{H}$) as a non-identity map.