1)
original
$$ \begin{bmatrix} 1 & 1 & 1 & 6 \\ 2 & 8 & 4 & 22\\ 1 & 1 & 7 & 18\\ \end{bmatrix} $$ reduced echelon form $$ \begin{bmatrix} 1 & 0 & 0 & 3.0 \\ 0 & 1 & 0 & 1.0\\ 0 & 0 & 1 & 2.0\\ \end{bmatrix} $$
Where I am stuck:
1) I did $-2R_1 + R_2$
2) I did $- R_1 - R_3 $
3) Reduce row $2$ and I get $$ \begin{bmatrix} 1 & 1 & 1 & 6.0 \\ 0 & 1 & 1/3 & 5/3\\ 0 & 0 & 1 & 2\\ \end{bmatrix} $$
On
Your first step is correct: $$-2R_1+R_2 \Rightarrow R_2.$$ Your second step should be: $$-R_1+R_3 \Rightarrow R_3.$$ Your third step should be: $$\frac{1}{6}R_2 \Rightarrow R_2$$ $$\frac{1}{6}R_3 \Rightarrow R_3.$$ The resulting form you wrote is correct. Next steps are: $$-\frac{1}{3}R_3+R_2 \Rightarrow R_2$$ $$-R_3+R_1 \Rightarrow R_1$$ $$-R_2+R_1 \Rightarrow R_1.$$
Your matrices are row equivalent so far and you have obtained a row echelon form.
To obtain the reduced row echlon form.
Perform $$R_2-\frac13R_3, R_1-R_3, R_1-R_2$$
Remark:
I use the convention of $R_i+cR_j$ to mean add $c$ times $R_j$ to $R_i$ and update $R_i$.