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).