I'm wondering if I have a valid answer to this.
It is exactly (e) of the following:
I first state that the two vector spaces are isomorphic because they have equal dimension. I then define a linear transform $S: \mathbb R^4 / Ker \ T \to \mathbb R^2$ by
$$S \begin{pmatrix} 0 \\ a \\ b \\ 0 \\ \end{pmatrix} \to \begin{pmatrix} a \\ b \\ \end{pmatrix}: a,b \in \mathbb R$$
Where $\begin{pmatrix} 0 \\ a \\ b \\ 0 \\ \end{pmatrix}$ is an arbitrary element in the departure space.
If this mapping translates basis to basis, this is an isomorphism.
$$ S \begin{pmatrix} 0 \\ a \\ b \\ 0 \\ \end{pmatrix} \to \begin{pmatrix} a \\ b \\ \end{pmatrix} \in \mathbb R^2 \implies S\left(a \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} + b\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix}\right) = a \ \vec e_1 + b \vec \ e_2 $$ $$ \text{ as $S\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$ and $ S\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}$}$$
Is this a valid answer?
Your answer is not valid, since you don't even mention $F^4/\ker T$ in it. An explicit isomorphism would be$$\begin{array}{rccc}\Psi\colon&F^4/\ker T&\longrightarrow&F^2\\&\begin{pmatrix}a\\b\\c\\d\end{pmatrix}+\ker T&\mapsto&T\begin{pmatrix}a\\b\\c\\d\end{pmatrix}.\end{array}$$