You know that the linear transformation
$$ T: \mathbb{R}^4 \to \mathbb{R}^4: \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} \mapsto T \begin{bmatrix} x \\ y \\ z \\ w \end{bmatrix} $$
has vectors $Ker(T)=[(1,0,0,1);(−1,0,0,1)]$ as kernel.
The span of these two vectors is equal to $$\begin{bmatrix} \alpha \\ 0 \\ 0 \\ \beta \end{bmatrix}$$
so the first and fourth vectors of the standard basis get mapped to zero.
Now, for the matrix of a linear transformation with respect to the standard basis, the first column is the image of the first basis vector, the second column is the image of the second basis vector and so on. Therefore, the first and fourth column of your $T$ must be zero, as the first and fourth basis vector lie in the kernel.
Expanding on this idea, it now suffices to choose the second column and third column as the given two vectors in the image $Im(T)=[(1,−1,0,2);(0,1,−1,0)]$, as this ensure that there is a domain vector which gets mapped to them.
So in conclusion, $T$ is (non-uniquely!)
$$ T = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 2 & 0 & 0 \end{bmatrix}.$$