What is the transformation matrix of the linear mapping?

$ f:M_{2.2}(\mathbb{R}) \rightarrow \mathbb{R^3}, $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto \begin{pmatrix} 2a - 4b \\ -6d \\ 8a-16b+2d \end{pmatrix}\beta =\big (\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} ,\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \big), \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \big), \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \big)$ $\beta'= ( e_{2}, 2e_{1}, -e_{3})$ of $\mathbb{R^{3}}$ ( these are the standard unit vectors of $\mathbb{R^{3}}$)

What is the transformation matrix $M_{\beta'}^{\beta}(f)= ?$

Well my thought is that

$f(\left( \begin{array}{c} 0\\ 1\\ 0 \end{array} \right)) =\left( \begin{array}{c} -4\\ 0\\ -16 \end{array} \right) = \square \left( \begin{array}{c} 1\\ 0\\ 8 \end{array} \right) +$$\square \left( \begin{array}{c} -4\\ 0\\ 2 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ 0\\ 0 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ -6\\ 2 \end{array} \right)$

$f(\left( \begin{array}{c} 2\\ 0\\ 0 \end{array} \right)) =\left( \begin{array}{c} 4\\ 0\\ 16 \end{array} \right) = \square \left( \begin{array}{c} 1\\ 0\\ 8 \end{array} \right) +$$\square \left( \begin{array}{c} -4\\ 0\\ 2 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ 0\\ 0 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ -6\\ 2 \end{array} \right) $

$f(\left( \begin{array}{c} 0\\ 0\\ -1 \end{array} \right)) =\left( \begin{array}{c} 0\\ 6\\ -2 \end{array} \right) = \square \left( \begin{array}{c} 1\\ 0\\ 8 \end{array} \right) +$$\square \left( \begin{array}{c} -4\\ 0\\ 2 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ 0\\ 0 \end{array} \right) +$ $\square \left( \begin{array}{c} 0\\ -6\\ 2 \end{array} \right) $

And from the linear combinations of the $squares$ I can fill the transformation matrix in the columns and in this way I will get the transformation matrix. Is my idea correct? And how does the transformation matrix look like? Thank you in advance

Your idea is not correct, since the domain of $f$ is $\Bbb R^{2\times2}$; so it makes no sense to talk about, say$$f\left(\begin{bmatrix}0\\1\\0\end{bmatrix}\right).$$

Note that\begin{align}f\left(\begin{bmatrix}1&0\\0&0\end{bmatrix}\right)&=\begin{bmatrix}2\\0\\8\end{bmatrix}\\&=0\times e_2+1\times(2e_1)-8(-e_3).\end{align}Therefore, the entries of the first column of the matrix that you're after are $0$, $1$, and $-8$.

Can you take it from here?