We have $\phi\colon \mathbb{R}^4 \to \mathbb{R}^4, \phi(x_1,x_2,x_3,x_4)= (x_1-2x_3,-2x_3-x_4,x_1+2x_3+2x_4,x_1+x_4)$. We want to find the matrix of this transformation with respect to the following basis of $\mathbb{R}^4$:
\begin{align} \alpha_1&=(1,1,0,0)\\ \alpha_2&=(0,1,1,0)\\ \alpha_3&=(0,0,1,1)\\ \alpha_4&=(0,0,0,1)\end{align}
First, we can write down the transformation as $$ A_\phi = \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 0 & -2 & -1 \\ 1 & 0 & 2 & 2 \\ 1 & 0 & 0 & 1\\ \end{bmatrix} $$ and simplify it to the reduced row echelon form $$ \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & {1 \over 2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{bmatrix} $$ From this, we immediately see $\dim\ker\phi = 2$ and, by extension, $$\dim\operatorname{im} \phi = \dim\mathbb{R}^4 - \dim\ker \phi = 4 - 2 = 2.$$ So, we can choose whichever two linearly independent column vectors from $A_\phi$ and they will be a basis of $\operatorname{im}\phi$.
Now, how do we proceed from here to find the matrix of the transformation with respect to the given basis?