I think we can consider this question in this way.
The matrix which does the rotation is then
$$
T =
\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}
$$
Then we do the shear transform. In the rotated system, the coordinates of two base vectors become $\hat{i}=\begin{bmatrix}1\\-1\end{bmatrix}$ and $\hat{j}=\begin{bmatrix}0\\1\end{bmatrix}$. So the transform is
$$
S =
\begin{bmatrix}
1 & 0\\
-1 & 1
\end{bmatrix}
$$
But we are in the transformed system. If we see this shear in the original system, it would be like
$$
T\cdot{S} =
\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0\\
-1 & 1
\end{bmatrix}
=
\begin{bmatrix}
1 & -1\\
1 & 0
\end{bmatrix}
$$
which is the composition of two transforms.
Hope this will help.
Update:
When we talk about a vector, say, $v = \begin{bmatrix}3\\4\end{bmatrix}$, we talk about it in the standard basis. I.e., $v = 3\hat{i} + 4\hat{j}$ or $v=\begin{bmatrix}\hat{i}&\hat{j}\end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix}$.
First, we do the rotation.
Base vectors change from $\hat{i}$, $\hat{j}$ to $\hat{i}' = \hat{j}$, $\hat{j}' = -\hat{i}$.
$$
T =
\begin{bmatrix}
\hat{i}' & \hat{j}'
\end{bmatrix}
=
\begin{bmatrix}
\hat{j} & -\hat{i}
\end{bmatrix}
$$
Then we do the shear transform.
Here for base vectors, $\hat{i}'' = \hat{i}' - \hat{j}'$, $\hat{j}'' = \hat{j}'$
$$
S =
\begin{bmatrix}
\hat{i}'' & \hat{j}''
\end{bmatrix}
=
\begin{bmatrix}
\hat{i}' - \hat{j'} & \hat{j}'
\end{bmatrix}
$$
But as you can see our base vectors are based on $\hat{i}'$ and $\hat{j}'$ instead of $\hat{i}$ and $\hat{j}$. So this only describes the shear transform in terms of the rotated system. But we know the relationship between them from the rotation transform. So we can get the transform
$$
T'=
\begin{bmatrix}
\hat{i} + \hat{j} & -\hat{i}
\end{bmatrix}
$$
which describes a rotation and a shear directly.
For injective: assume $\Phi(T)=0$ then
$[T]^\gamma_\beta=0 \Rightarrow T=0$.
So $N(\Phi)=\{0\}$ $\Rightarrow$ $\Phi$ is injective.
For surjective: Take a matrix, define a linear transformation then how are these two the linear transformation and matrix are defined? The linear transformation must be such that the matrix of the linear transformation relative to the basis $\beta$ and $\gamma$ is the matrix that you started with.
Can you show that $\Phi$ is linear transformation?
Best Answer
I hope you understand what they are trying to achieve: we are looking for a linear transformation $T$ which reflects points of $\Bbb R^2$ on the line $y=2x$. We will completely know a linear transformation, if we know it's action on a basis of $\Bbb R^2$. You can check independently, that the vectors
$$\begin{bmatrix} 1 \\2 \end{bmatrix}, \begin{bmatrix}-2 \\ 1 \end{bmatrix} $$
form a basis for $\Bbb R^2. $So we will know $T$ if we know it's action on this basis. Call this basis $\beta'$. Why are we choosing this basis? This is because it's easy to see $T$'s action on $\beta'$. Since $T$ reflects points on the line $y = 2x$, you can easily check that $$T \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix}1 \\2 \end{bmatrix}$$
this is because the point $(1,2)$ lies on the line of reflection. Similarly using geometric arguments, you can show that $T(-2,1) = (2,-1)$. Therefore the matrix of $T$ with respect to the basis $\beta'$ is $$\begin{bmatrix}1 & -2 \\2 & 1 \end{bmatrix}$$
Since we are used to dealing with vectors in the usual canonical basis, we simply do a change of basis and find the corresponding matrix of $T$ with respect to the usual basis.