So if I have a function $f: M_{2,2} → ℝ^2$ where $\begin{bmatrix}a&b\\c&d\end{bmatrix} → \begin{bmatrix}a&-b\\c&-d\end{bmatrix}$. How would I find $[f]_α^β$ if $α = ${$\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}$} and β = {$\vec e_1, \vec e_2$}.
The answer says it's $\begin{bmatrix}1&-1&0&0\\0&0&1&-1\end{bmatrix}$.
So how do I get to that answer but also in general how would I find the answer if I had to find $[f]_α^β$
Let $V$ be an $n$-dimensional vector space over $\mathbb{F} $ and $W$ be an $m$-dimensional vector space over $\mathbb{F}$. Given a linear map $T \colon V \rightarrow W$, a basis $\beta = (v_1, \dots, v_n)$ for $V$ and $\alpha = (w_1, \dots, w_m)$ for $W$, the matrix $[T]^{\beta}_{\alpha} \in M_{m \times n}(\mathbb{F})$ representing $T$ with respect to the bases $\alpha$ and $\beta$ is given by
$$ [T]_{\alpha}^{\beta} = \begin{pmatrix} [Tv_1]_{\alpha} & \dots & [Tv_n]_{\alpha} \end{pmatrix}. $$
In other words, you need to apply $T$ to each basis element of $v_i$ and write the result $Tv_i$ as a linear combination in terms of the basis elements $\beta$. The linear combination will then be the $i$-th column of the matrix $[T]_{\alpha}^{\beta}$.
For example, in your case
$$ f \left( \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \right) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = 1 \cdot \vec{e_1} + 0 \cdot \vec{e_2} $$
so the matrix $[f]_{\alpha}^{\beta}$ looks like
$$ [f]_{\alpha}^{\beta} = \begin{pmatrix} 1 & ? & ? & ? \\ 0 & ? & ? & ? \end{pmatrix}. $$
Make sure you understand how the other columns of $[f]_{\alpha}^{\beta}$ are obtained.