Hint:
Notice that each vector in $\beta$ is a linear combination of vectors in $\beta_s$: this defines what $\beta_s$ is in terms of $\beta$. So, to find $Q$, simply write out the matrix whose columns represent these linear combinations. The rest is as you say.
The $i$-th column of the matrix $L_\alpha^\beta$ contains the image of $\alpha_i$ with respect to the basis $\beta$. Since the second column is the zero vector, it is the image of an element of the null space (or kernel).
Based on your edit, I believe you were already looking in this direction. I'd take $(0,1,-2,2)$ but yours works too. Since you want its image in the second column, you take this vector as $\alpha_2$. Extend $\alpha$ to a complete basis of $\mathbb{R^4}$, e.g. by adding standard basis vectors:
$$\alpha = \left\{
\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \\ -2 \\ 2 \end{pmatrix},
\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}
\right\}$$
Now the trick to the simple form of $L_\alpha^\beta$ is to choose the basis vectors of $\beta$ carefully: use the images of the $\alpha$'s! That is, pick: $\beta_j = L(\alpha_j)$ for $j=1,3,4$. Not for $j=2$, because by our choice of $\alpha_2$, we have $L(\alpha_2) = 0$. Extend $\beta$ to a basis of $\mathbb{R^4}$ by picking, for example, $\beta_2 = (0,0,1,0)^T$.
For $\beta_1$, we find the image of $\alpha_1$:
$$\beta_1 = L(\alpha_1) = L \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} =\begin{pmatrix} 2 \\ 0 \\ -2 \\ 0 \end{pmatrix} $$
Repeat this for $\beta_3$ and $\beta_4$ to get:
$$\beta = \left\{
\begin{pmatrix} 2 \\ 0 \\ -2 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 2 \\ 4 \\ 0 \end{pmatrix},
\begin{pmatrix} 0 \\ 0 \\ 0 \\ 5 \end{pmatrix}
\right\}$$
Notice that other choices are possible: you fix $\alpha_2$ and then complete $\alpha$ to a basis by adding three linearly independent vectors. You then choose the $\beta$'s as the images of these three basis vectors you added to $\alpha$ to obtain the given, simple form of the transformation matrix. These three will span the image of $L$ and you simply extend $\beta$ to a basis by adding any linearly independent vector.
Best Answer
The matrix $T$, with respect to the basis $v_1= 1 + x + x^2$, $v_2= x + x^2$, and $v_3=x^2$ can be computed directly as follows
$T(v_1)=T(1+x+x^2)=(1-1+1) + (1+1)x + (2-1)x^2=1 + 2x + x^2= v_1+ v_2 - v_3$,
$T(v_2)=T(x+x^2)= 2x + 2x^2 = 2v_2$,
$T(v_3)=T(x^2)= 1 + x = v_1-v_3$
Therefore, the matrix $T=(a_{ij})$ is the matrix which solves the equation
$\begin{pmatrix} v_1 & v_2 & v_3 \end{pmatrix} T = \begin{pmatrix} v_1 + v_2 - v_3 & 2v_2 & v_1-v_3 \end{pmatrix}$. I'll let you solve explicitly for the values of $T$.