ZettaSuro has given you the explanation but let me add a bit more to help you
understand the problem deeply. Consider the following example of a transformation.
$
T: R^{2}\rightarrow R^{2} \mbox{ given by } T(x_{1},x_{2})=2x_{1}+3x_{2}
$
All this transformation is doing is taking a vector in the plane and scaling it in each respective direction i.e $x$ and $y$. Now there is one idea that needs to be emphasized behind these types of transformations.
Any linear transformation can be represented by a matrix or a matrix is really a linear transformation from one space to another
If we change our example to the one below
$
T: R^{2}\rightarrow R^{2} \mbox{ given by } T(x_{1},x_{2})=2x_{1}^{2}+3x_{2}
$
then we no longer have a linear transformation(check for yourself) and there is no representation of the transformation using a matrix.
So how do you find the matrix that corresponds to a linear transformation? All you need to do is consider what the transformation does to the basis vectors. Let's see the first example above. In this case the basis vectors are $\begin{bmatrix}0\\1 \end{bmatrix}$ and $\begin{bmatrix}1\\0 \end{bmatrix}$.
$
T \begin{bmatrix}0\\1 \end{bmatrix}=\begin{bmatrix}0\\3 \end{bmatrix}
$ and $
T \begin{bmatrix}1\\0 \end{bmatrix}=\begin{bmatrix}2\\0 \end{bmatrix}
$
Now, here is the key. Consider any other vector $\begin{bmatrix}a\\b \end{bmatrix}$ on the plane and apply the transformation
$$
T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=T\left(a\begin{bmatrix}1\\0 \end{bmatrix}
+b\begin{bmatrix}0\\1 \end{bmatrix} \right)\\
T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=aT\begin{bmatrix}1\\0 \end{bmatrix}
+bT\begin{bmatrix}0\\1 \end{bmatrix} \\
T\left(\begin{bmatrix}a\\b \end{bmatrix}\right)=\begin{bmatrix} T\begin{bmatrix}1\\0 \end{bmatrix}& T\begin{bmatrix}0\\1 \end{bmatrix}
\end{bmatrix} \begin{bmatrix} a\\ \\b \end{bmatrix}
$$
Now we can just read the matrix from the calculation and the matrix is
\begin{bmatrix}
2 & 0\\
0 & 3\\
\end{bmatrix}
You can apply this for your case and get the matrix below
\begin{bmatrix}
1 & 1 & 0\\
0 & -2 & 3\\
\end{bmatrix}
Note that
$$\begin{bmatrix}
1 & -1 & 1\\
0 & 1 & -2\\
0 & 0 & 1
\end{bmatrix} \cdot \begin{bmatrix} a_0 \\ a_1 \\ a_2 \end{bmatrix} =
\begin{bmatrix} a_0 - a_1 + a_2 \\ a_1 - 2a_2 \\ a_2 \end{bmatrix}.$$
At the same time,
\begin{align}
a_0 + a_1(x-1) + a_2(x-1)^2 &= a_0 + a_1x - a_1 + a_2x^2 - 2a_2x + a_2 \\
&= (a_0 - a_1 + a_2) + (a_1 - 2a_2)x + a_2x^2.
\end{align}
I hope it is more clear this way.
Best Answer
Perhaps better notation would be $T\left( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \right) = (a+b-c-d)t^2 + (c+d)t + (a+b) $.
$T$ is a function. The matrix $\begin{bmatrix} a & b \\ c & d\end{bmatrix}$ is the input of $T$. The output of $T$ is the polynomial $(a+b-c-d)t^2 + (c+d)t + (a+b)$.