We do want to use a kind of Gaussian elimination, but you have to be careful since you should not multiply a row by anything other than $1$ and $-1$, and you should not add non-integer multiples of one row to another row. So we can get started simply enough:
$$\begin{align*}
\left(\begin{array}{rrrr}
2 & 4 & 6 & -8 \\
1 & 3 & 2 & -1 \\
1 & 1 & 4 & -1 \\
1 & 1 & 2 & 5
\end{array}\right)
&\to
\left(\begin{array}{rrrr}
1 & 1& 2 & 5\\
1 & 3 & 2 & -1\\
1 & 1 & 4 & -1\\
2 & 4 & 6 & -8
\end{array}\right)
&&\to \left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 2 & 2 & -18
\end{array}\right)\\
&\to \left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 2 & -12
\end{array}\right)
&&\to\left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 0 & -6
\end{array}\right)\\
&\to\left(\begin{array}{rrrr}
1 & 1 & 2 & 5\\
0 & 2 & 0 & -6\\
0 & 0 & 2 & -6\\
0 & 0 & 0 & 6
\end{array}\right).
\end{align*}$$
This uses only elementary row operations.
From this we see that the relations on your group are equivalent to:
$$\begin{array}{rcccccccl}
r_1&+&r_2&+&2r_3&+&5r_4 &= & 0\\
& &2r_2 & & & - & 6r_4 &=& 0\\
& & & & 2r_3 & - &6r_4 & = & 0\\
& & & & & &6r_4 & = & 0
\end{array}$$
These elementary row operations replace the relations on our original set of generators with a new set of relations which are equivalent to the original, in the sense that if the generators satisfy these relations, then they satisfy the original relations and vice-versa.
We can now use elementary column operations, which also correspond to performing certain base changes. For example, subtracting five times the first column from the fourth column is equivalent to replacing $r_1$ with $r_1-5r_4$, which does not change the subgroup generated by $r_1,r_2,r_3,r_4$. Etc. Performing those elementary column operations, since the $(1,1)$ entry is the gcd of the entries on the first row (and similarly for the rest of the rows), we can eliminate all nondiagonal entries and end up with the diagonal matrix
$$\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0\\
0 & 0 & 2 & 0\\
0 & 0 & 0 & 6
\end{array}\right),$$
from which you can read off the structure of the group in question.
The column operations corresponds to replacing our set of generators with a new set that generates the same group. For example, the operations we performed with the first column, subtracting the first column from the second, twice the first column from the third, and five times the first column from the fourth, correspond to replacing the original generator $r_1$ with the generator $r_1-r_2-r_3-5r_4$. This does not change the subgroup, because $$\langle r_1,r_2,r_3,r_4\rangle = \langle r_1-r_2-r_3-5r_4,r_2,r_3,r_4\rangle.$$
Similarly with the other operations. In the end we will have an abelian group generated by elements $a,b,c,d$, where, in terms of the original generators, we have
$$\begin{align*}
a & = r_1-r_2-r_3-5r_4\\
b &= r_2-3r_4\\
c &= r_3-3r_4\\
d &= r_4,
\end{align*}$$
which yields the relations
$$\begin{align*}
a&=0\\
2b&=0\\
2c&=0\\
6d&=0
\end{align*}$$
from which we can just read off the abelian group structure as well.
The tool for this is the Smith Normal Form, a kind of Gaussian elimination for PID.
For the example $G = \langle s,t,u,v \mid s^{4}t^{2}u^{10}v^{6} = s^{8}t^{4}u^{8}v^{10} = s^{6}t^{2}u^{9}v^{8} = e_G \rangle$ written additively, the matrix is
$$
\left(
\begin{array}{cccc}
4 & 2 & 10 & 6 \\
8 & 4 & 8 & 10 \\
6 & 2 & 9 & 8 \\
\end{array}
\right)
$$
whose Smith Normal Form is
$$
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 \\
0 & 0 & 2 & 0 \\
\end{array}
\right)
$$
as computed by Mathematica using the code mentioned here.
This means that $G$ is generated by four elements $g_1, g_2, g_3, g_4$ such that $1g_1=0$, $2g_2=0$, $2g_3=0$, and no restrictions on $g_4$, and so $G \cong C_1 \times C_2 \times C_2 \times C_{\infty} \cong C_2 \times C_2 \times C_{\infty}$.
$g_1, g_2, g_3, g_4$ are obtained from $s,t,u,v$ by applying the matrices $P$ and $Q$ such that $PAQ$ is the diagonal matrix above.
All this is explained in several books. One is
Finitely Generated Abelian Groups and Similarity of Matrices over a Field by Christopher Norman. A shorter account appears in Jacobson's Basic Algebra I.
Best Answer
For each matrix $A\in \mathrm{Mat}_{4}(\mathbb{Z})$, the map $$\begin{array}{rcl} \mathbb{Z}^4 & \to & \mathbb{Z}^4\\ \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]&\mapsto & A\cdot\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\end{array}$$ is a group homomorphism of Abelian groups. It is moreover a group isomorphism if and only if $A\in \mathrm{GL}_4(\mathbb{Z})$, which corresponds to $\det(A)=\pm 1$.
You are trying to get the subgroup of $\mathbb{Z}^4$ given by $$\left\{\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\in \mathbb{Z}^4\ \left|\ R\cdot\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]=0\right\}\right.$$ which is the kernel of the group homomorphism given by $$\begin{array}{rcl} \mathbb{Z}^4 & \to & \mathbb{Z}^4\\ \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]&\mapsto & R\cdot \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\end{array}$$ Applying an automorphism of $\mathbb{Z}^4$ at the target does not change the kernel. Applying an automorphism at the source changes the kernel, but only by an isomorphism, so if you want to compute the isomorphism class of your group, you can multiply $R$ at the right and left by elements of $\mathrm{GL}(4,\mathbb{Z})$. In particular, you can add any multiple of any row of $R$ to another one, and the same for columns. You can also exchanges lines or columns and multiply one line or one column by $-1$.
At the end you get a diagonal matrix which is called Smith normal form. There is a short algorithm but you can do it easily by hand. For instance, here subtracting the first column to the others gives $$\left[\begin{array}{cccc}2&2&2&2\\4&4&8&5\\6&12&12&8\\4&10&8&6\end{array}\right]\to \left[\begin{array}{cccc}2&0&0&0\\4&0&4&1\\6&6&6&2\\4&6&4&2\end{array}\right].$$ You can then add multiples of the first line to the others and get $$\left[\begin{array}{rrrr}2&0&0&0\\4&0&4&1\\6&6&6&2\\4&6&4&2\end{array}\right]\to \left[\begin{array}{rrrr}2&0&0&0\\0&0&4&1\\0&6&6&2\\0&6&4&2\end{array}\right].$$ Then you exchange the second and fourth column and remove four times this one to the second to get $$\left[\begin{array}{rrrr}2&0&0&0\\0&0&4&1\\0&6&6&2\\0&6&4&2\end{array}\right]\to \left[\begin{array}{rrrr}2&0&0&0\\0&1&4&0\\0&2&6&6\\0&2&4&6\end{array}\right]\to \left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&2&-2&6\\0&2&-4&6\end{array}\right].$$ The last steps are the following $$ \left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&2&-2&6\\0&2&-4&6\end{array}\right]\to\left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&-2&6\\0&0&-4&6\end{array}\right]\to\left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&-2&6\\0&0&0&-6\end{array}\right]\to\left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&-2&0\\0&0&0&-6\end{array}\right].$$ Hence your group is isomorphic to the quotient of $\mathbb{Z}^4$ by the kernel of $$\begin{array}{rcl} \mathbb{Z}^4 & \to & \mathbb{Z}^4\\ \left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]&\mapsto & \left[\begin{array}{rrrr}2&0&0&0\\0&1&0&0\\0&0&2&0\\0&0&0&6\end{array}\right]\cdot\left[ \begin{array}{c} a_1\\ a_2 \\ a_3 \\ a_4\end{array}\right]\end{array}$$ and is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/6\mathbb{Z}$.
If you want the generators of your group you can also take them back by the isomorphisms. You can also first do operations on the rows, so that you do not change the kernel, and obtain a triangular matrix.