Yes precisely. The number of variables you have (call this $m$) tells you the dimension of the vector space you are in. As you constrain more of the variables, the possible solution set reduces. If you have $n$ free variables at the end, then this is an $n$ dimensional surface in your $m$ dimensional space. ($n\le m$ always)
So when $n=2$, i.e. you have $2$ free variables, you will have a plane of solutions. If, say, $n=3$,$m=5$ then you have a $3D$ plane (referred to as a hyperplane) in a $5D$ space (this isn't really possible for us to visualise however).
In these sort of cases, the solution is a line or plane or something similar, embedded within the input space. When describing this solution we can choose any variable that is not orthogonal to it as a free variable, and then once we have enough to fully describe a projection of the solution with the same number of dimensions, we can say any other variables we would need to get back to the solution set have you are "basic", forced by the selection of free variables to take on a particular value.
In case 1, the solution is $x = (\frac{109}{42}, \frac{1}{21}, -\frac{55}{42}) + (2,1,4)t$, and since none of $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ are orthogonal to $2,1,4$, we can pick any one of $x_1$, $x_2$, and $x_3$ to be the free variable.
In case 2, the solution can be described as the line $x = (0,5) + (1,0)t$, where it is obvious that $(0,1)$ is orthogonal to the solution line, and so $x_2$ cannot be considered free.
I've done a bit of searching to figure out how to detect this in the RREF. here's an example.
$$
\begin{pmatrix}
1\\
4
\end{pmatrix}= \begin{pmatrix}
4& -6 & 3\\
-2 & 3 & 4
\end{pmatrix}\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}
$$
This has RREF
$$
\left(\begin{array}{ccc|c}
1 & -\frac{3}{2} & 0 & -\frac{4}{11}\\
0 & 0 & 1 & \frac{9}{11}
\end{array}\right)
$$
And it turns out that $x_3 = \frac{9}{11}$ all the time. The solution set is $(-\frac{2}{11}, \frac{4}{33}, \frac{9}{11}) + (2,3,0)t$; this is clearly orthogonal to $(0,0,1)$ and so $x_3$ is fixed.
This provides guidance to the answer The second row of the RREF contains only columns from $x_3$ and the solution vector, so this one must be fixed; I can select from $x_1$ and $x_2$ to be free because they share a row.
Applying this wisdom to case 1: selecting one of $x_1$ or $x_2$ to be free fixes $x_3$ by the action of one of the rows, which then (since everything else in the other row is fixed) fixes the other one, and picking $x_3$ to be free fixes $x_1$ and $x_2$ by the respective actions of the two rows in themselves.
Applying similar wisdom to case 2: $x_1$ gets to be free because it affects absolutely nothing; $x_2$ does not because it is surrounded by already-fixed results.
So it looks like it goes something like this: if a variable is in a position in one of the rows where it can affect only variables that are either fixed or already chosen as free, then it must be fixed. Otherwise, you may choose it to be free.
Best Answer
$x+y+w=0$
$x-z+2w=0$
The first equation leads to $x = -y-w$ and the second leads to $x = z-2w$. Then we have $-y-w = z-2w$, so $y = -z+w$
Then $x = z-2w$ and $y = -z+w$. We have two variables $z,w$ that are free, as expected.
So you have a vector $(x,y,z,w)$ and want to write it as a combination of basis vectors. As the equations showed: $(x,y,z,w) = (z-2w,-z+w,z,w) = (1,-1,1,0)z + (-2,1,0,1)w$, hence a basis would be $\{(1,-1,1,0),(-2,1,0,1)\}$