There are many many reasons why kernels are important to group theory, but here's just one way of appreciating the kernel in a fairly isolated context.
If we zoom out a bit, any set-function $f: A \to B$ (here $A$ and $B$ are simply sets) naturally partitions $A$ into equivalence classes, and for $a \in A$, the equivalence class of $a$ is given by
$$[a] = \{a' \in A : f(a') = f(a)\};$$
the set of all elements of $A$ that get mapped to the same thing as $a$. The same logic applies if $f : G \to H$ isn't just a set function, but a homomorphism of groups.
With the equivalence class notation, the kernel of $f$ is simply the equivalence class of the identity $1_G$ of $G$,
$$\ker f = [e_G],$$
since any homomorphism $f: G \to H$ always sends the identity $e_G$ of $G$ to the identity $e_H$ of $H$. What can we say about arbitrary $g, g' \in G$ such that $f(g) = f(g')$? That is, what can we say about the equivalence class $[g]$ for any $g \in G$?
Claim: For a homomorphism $f: G \to H$ and $g \in G$, we have $f(g) = f(g')$ if and only if there exists some $k \in \ker f$ such that $gk = g'$; that is, $g$ and $g'$ differ by a multiple of something in the kernel of $f$. In particular, $[g] = \{gk: k \in \ker f\}$, and has size $|\ker f|$.
($\Longrightarrow$) Supposing $f(g) = f(g')$, note that there exists a unique $g^* = g^{-1}g' \in G$ such that $gg^* = g'$. Then
$$f(g) = f(g') = f(gg^*) = f(g)f(g^*),$$
and left-multiplication by $f(g)^{-1}$ shows that $e_H = f(g^*)$, hence $g^* \in \ker f$.
($\Longleftarrow$) Homework.
If you've ever heard homomorphisms described as functions that "respect" the group operation(s), the size of the kernel is a measure of just how "respectful" a given homomorphism is! A large kernel means that more of the structure of the group $G$ is "ignored" when transported to the group $H$.
Edit:
For "respectful", imagine two situations, considering $S_3$, the symmetric group of degree $3$. There's a sign homomorphism $\operatorname{sgn} : S_3 \to \{-1, 1\} = C_2$ to the multiplicative group $C_2$ sending each permutation to its sign. Its kernel is the alternating group $A_3 = \{1, (123), (132)\}$ of "even" permutations, and in the image $C_2$, almost all of the structure of $S_3$ is ignored; we forget everything but whether a permutation is even or odd.
On the other hand, we have a "copy" of $S_3$ as a subgroup of $S_4$ if we consider all permutations of $S_4$ that leave $4$ fixed. This leads to an "inclusion" homomorphism $\iota: S_3 \to S_4$, sending each permutation to its "copy" in $S_4$. This inclusion homomorphism has only the identity of $S_3$ in its kernel, and is considerably more "respectful" than the sign homomorphism; every bit of information about $S_3$ shows up in the image $\iota(S_3)$.
In order that an upper triangular matrix
$$\begin{pmatrix} a & b & c \\ 0 & d& e \\ 0 &0 & f\end{pmatrix}$$
be invertible, it is necessarily and sufficient that the determinant $adf$ be nonzero. This means that $b,c,e$ can be anything you want, while $a, d, f$ cannot be zero. This means there are
$$3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 \cdot 2 = 216 $$
such matrices.
The subgroup $D$ of diagonal invertible matrices is as far from being normal in $G$ as you can get (if $x \in G$, but not in $D$, then $xDx^{-1} \neq D$).
However, if you multiply two upper triangular matrices $x, y \in G$, notice that the entries on the diagonal of $xy$ are obtained by multiplying the corresponding entries on the diagonal of $x$ and $y$. This implies that
$$N = \{ \begin{pmatrix} 1 & b & c \\ 0 & 1& e \\ 0 &0 & 1\end{pmatrix} : b, c, e \in \mathbb{F}_3 \}$$ is a normal subgroup of $G$.
Actually, $G$ is the semidirect product of $N$ and $D$.
Best Answer
You might imagine it as the group of upper triangular matrices with the top-right entry marked as unknown or irrelevant, $$\begin{pmatrix}x&y&?\\0&z&u\\0&0&v\end{pmatrix} $$ This is fine because when multiplying two such matrices, the top-right entries are needed only for computing the top-right entry: $$\begin{pmatrix}x&y&?\\0&z&u\\0&0&v\end{pmatrix}\begin{pmatrix}x'&y'&?\\0&z'&u'\\0&0&v'\end{pmatrix}=\begin{pmatrix}xx'&xy'+yz'&?\\0&zz'&zu'+uv'\\0&0&vv'\end{pmatrix} $$
By removing all decoration, this becomes a group structure on the set $$G:=\{\,(x,y,z,u,v,D)\mid yzvD-1=0\,\}$$ and with multiplication rule $$(x,y,z,u,v,D)\cdot(x',y',z',u',v',D')=(xx',xy'+yz',zz',zu'+uv',vv',DD').$$ But I suppose that this explicit rule looks a bit unintuitive, compared to the matrix with irrelevant entry.
As a sidenote: Because the set $G$ and the group operation are defined in terms of polynomials, this is an algebraic group