How do I complete solving this homogeneous system of linear equations using Gaussian elimination? $${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }=0\\ { x }_{ 1 }+{ x }_{ 2 }+2{ x }_{ 3 }-{ x }_{ 4 }=0\\ 2{ x }_{ 1 }+2{ x }_{ 2 }+{ x }_{ 3 }+{ x }_{ 4 }=0\\ -{ x }_{ 1 }-{ x }_{ 2 }+{ x }_{ 3 }-2{ x }_{ 4 }=0\\ $$
Steps I took:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    1 & 1 & 2 & -1 &  0 \\
    2 & 2 & 1 & 1  & 0 \\
   -1 & -1 & 1 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 2 }={ 2R }_{ 2 }-{ R }_{ 3 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    2 & 2 & 1 & 1  & 0 \\
   -1 & -1 & 1 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 3 }={ R }_{ 3 }+{ 2R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 3 & -3  & 0 \\
   -1 & -1 & 1 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 4 }={ R }_{ 1 }+{ R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 3 & -3  & 0 \\
    0 & 0 & 2 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 3 }={ R }_{ 3 }-{ R }_{ 2 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 0 &  0  & 0 \\
    0 & 0 & 2 & -2 & 0
  \end{array}\right]$$
$${ R }_{ 2 }=2{ R }_{ 2 }-3{ R }_{ 4 }\\ \qquad \rightarrow $$
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 1 & 0 &  0 \\
    0 & 0 & 3 & -3 &  0 \\
    0 & 0 & 0 &  0  & 0 \\
    0 & 0 & 0 &  0 & 0
  \end{array}\right]$$
At this point I am stuck. I am having a lot of trouble figuring out what I need to do once I complete all the necessary row operations on my matrix. I would like to be guided through the process as of this point. 
Please explain it to me in layman terms. I am hoping to use what I learn from being guided through the final steps of this problem in order to understand how to solve ones without any hand holding. 
I'm also really frustrated with the fact that I am doing these operations like a computer without actually understanding what is going on with the "independent" and "dependent" variables.

You end up with 
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1&  0 \\
    0 & 0 & 1 & -1&  0 \\
  \end{array}\right].$$
All you have to do is paint in just straightening the tableau, i.e., create an upper triangle matrix whose diagonal elements are $\pm1$ in the following way. The first row of your reduced matrix has the first $1$  at position $1$ so that would be the first row, the second row's position of the leading $1$ is $3$ so that becomes the third row; just leave space for row $2$ and $4$:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1 &  0 \\
\strut &&&&\\
    0 & 0 & 1& -1&  0 \\
\strut &&&&
  \end{array}\right].$$
Now fill the empty rows with zeroes except for the row number's position: you insert a $-1$: the second row consists of zeroes except for the second positon, you fill in a $-1$ here:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1&  0 \\
0&-1&0&0&0\\
    0 & 0 & 1& -1&  0 \\
\strut &&&&
  \end{array}\right].$$
Row number four gets a $-1$ at position four:
$$\left[\begin{array}{rrrr|r}
    1 & 1 & 0& 1&  0 \\
0&-1&0&0&0\\
    0 & 0 & 1& -1&  0 \\
0&0&0&-1&0
  \end{array}\right].$$
Now the columns with the numbers of the inserted rows, namely column two 
$\left[\begin{array}{c}
    1 \\-1\\ 0 \\0
  \end{array}\right]$
and column four
$\left[\begin{array}{r}
    1\\ 0\\ -1\\ -1
  \end{array}\right]$
span the kernel of the matrix.  Cute, isn't it?
