What you have is an expression for every vector in the subspace in parametric form, with three parameters:
$$\begin{align*}
x_1 &= -2r + s\\
x_2 &=r\\
x_3 &=s\\
x_4 &=t
\end{align*}$$
with $r,s,t\in\mathbb{R}$, arbitrary.
To get a basis for the space, for each parameter, set that parameter equal to $1$ and the other parameters equal to $0$ to obtain a vector. Each parameter gives you a vector. So setting $r=1$ and $s=t=0$ gives you one vector; setting $s=1$ and $r=t=0$ gives you a second vector; setting $t=1$ and $r=s=0$ gives you a third.
Alternatively, you can try rewriting the parametric solution in vector form:
$$\left(\begin{array}{c}x_1\\x_2\\x_3\\x_4\end{array}\right) = \left(\begin{array}{c}-2r-s\\r\\s\\t\end{array}\right) = \left(\begin{array}{r}-2\\1\\0\\0\end{array}\right)r + \cdots$$
(I'll let you finish it up).
The null space of the matrix
$$\begin{bmatrix}
1 & 1 & -1 & -1 & 0 & 0 \\
1 & 1 & 0 & 0 & -1 & -1 \\
0 & 0 & 1 & 1 & -1 & -1 \\
\end{bmatrix}$$
is, by definition, the set of vectors $\mathbf{x}=(x_1,x_2,x_3,x_4,x_5,x_6)$ such that
\begin{align*}
x_1 + x_2 - x_3 - x_4 &= 0 \\
x_1 + x_2 - x_5 - x_6 &= 0 \\
x_3 + x_4 - x_5 - x_6 &= 0
\end{align*}
So, this is the vector space $V$ in question. The matrix has rank $2$, so the dimension of $V$ is $4$ by the Rank Nullity Theorem.
Thus, any $4$ linearly independent vectors in $V$ form a basis. A natural try is $$\{( 1, 1, 1, 1, 1, 1 ),(0, 0, 0, 0, 1, -1),( 0, 0, 1, -1, 0, 0 ),(1, -1, 0, 0, 0, 0)\}.$$ To prove that it is indeed a basis, we check that the matrix $$\begin{bmatrix}
1 & 1 & 1 & 1 & 1 & 1 \\
0 & 0 & 0 & 0 & 1 & -1 \\
0 & 0 & 1 & -1 & 0 & 0 \\
1 & -1 & 0 & 0 & 0 & 0 \\
\end{bmatrix}$$ has rank $4$ (e.g. by performing Gaussian elimination).
The column space of the transpose of the above matrix, i.e.
$$\begin{bmatrix}
1 & 0 & 0 & 1 \\
1 & 0 & 0 & -1 \\
1 & 0 & 1 & 0 \\
1 & 0 & -1 & 0 \\
1 & 1 & 0 & 0 \\
1 & -1 & 0 & 0 \\
\end{bmatrix},$$
will be $V$ (or, at least, will be isomorphic to $V$: all the vectors will be transposed); see Wikipedia.
Best Answer
Hint: Any $2$ additional vectors will do, as long as the resulting $4$ vectors form a linearly independent set. Many choices! I would go for a couple of very simple vectors, check for linear independence. Or check that you can express the standard basis vectors as linear combinations of your $4$ vectors.