The systems of linear equations $$x_1 - x_2 + 2x_3 - x_4= 6$$ $$x_1 - x_3 + x_4 = 4$$ $$2x_1 + x_2 + 3x_3 - 4x_4 = -2$$ $$-x_2 + x_3 - x_4 = 5$$ and $$x_1 - x_2 + 2x_3 - x_4= 1$$ $$x_1 - x_3 + x_4 = 1$$ $$2x_1 + x_2 + 3x_3 - 4x_4 = 2$$ $$-x_2 + x_3 - x_4 = -1$$ give \begin{bmatrix}1&-1&2&-1&|&6&1\\1&0&-1&1&|&4&1\\2&1&3&-4&|&-2&2\\0&-1&1&-1&|&5&-1\end{bmatrix} $$R_1 + (-1)R_2\rightarrow R_2$$ $$R_1 + (-12)R_3\rightarrow R_3$$ \begin{bmatrix}1&-1&2&-1&|&6&1\\0&-1&3&-2&|&2&0\\0&-3/2&1/2&1&|&7&0\\0&-1&1&-1&|&5&-1\end{bmatrix} $$R_2 + (-2/3)R_3\rightarrow R_3$$ $$R_2 + (-1)R_4\rightarrow R_4$$ \begin{bmatrix}1&-1&2&-1&|&6&1\\ 0&-1&3&-2&|&2&0\\ 0&0&8/3&-8/3&|&-8/3&0\\ 0&0&2&-1&|&3&1\end{bmatrix} $$(-1)R_2 \rightarrow R_2$$ $$(3/8)R_3 \rightarrow R_3$$ $$R_3 + (-8/6)R_4 \rightarrow R_4$$ \begin{bmatrix}1&-1&2&-1&|&6&1\\ 0&1&-3&2&|&-2&0\\ 0&0&1&-1&|&-1&0\\ 0&0&0 &-4/3&|&4/3&-4/3\end{bmatrix} $$(-3/4)R_4 \rightarrow R_4$$ \begin{bmatrix}1&-1&2&-1&|&6&1\\ 0&1&-3&2&|&-2&0\\ 0&0&1&-1&|&-1&0\\ 0&0&0&1&|&-1&1\end{bmatrix}
The problem is that I don't understand how $x_4$ can be both $-1$ and $1$, or how that applies to the rest of the equations.
You are solving two systems of linear equations at the same time. Since the two systems have the same coefficient matrix, this is possible by putting two columns next to the coefficient matrix, just like you did.
In the first system of equations, $x_4=-1$; and in the second system of equations, $x_4=1$. You should not mix the solutions of the two systems.