List the elements of $\operatorname{GL}(2, \mathbb{F}_2)$ and then look at the action of this group on the set of non-zero vectors in $\mathbb{F}^2_2$. Prove this allows you to identify the group $\operatorname{GL}(2, \mathbb{F}_2)$ with the symmetric group $S_3$.
So here are all the elements of $\operatorname{GL}(2, \mathbb{F}_2)$: $$\begin{pmatrix}1&0\\0&1\end{pmatrix},\quad \begin{pmatrix}0&1\\ 1&0\end{pmatrix},\quad \begin{pmatrix}1&1\\1&0\end{pmatrix},\quad \begin{pmatrix}1&1\\0&1\end{pmatrix},\quad \begin{pmatrix}1&0\\1&1\end{pmatrix},\quad \begin{pmatrix}0&1\\1&1\end{pmatrix}.$$
These are also all non-zero vectors in $\mathbb{F}^2_2$, additionally with $\begin{pmatrix}1&1\\1&1\end{pmatrix}$ since this is also a non-zero vector.
The group $\operatorname{GL}(2, \mathbb{F}_2)$ is closed under matrix multiplication since it is a group, so the action of this group on the non-zero vectors represented by these same elements result in elements in this group, so they permute. But with the non-zero vector $\begin{pmatrix}1&1\\1&1\end{pmatrix}$ of $\mathbb{F}^2_2$, if you multiply this with, say, the identity $\begin{pmatrix}1&0\\0&1\end{pmatrix}$, then you get back $\begin{pmatrix}1&1\\1&1\end{pmatrix}$ which is not a permutation since this is not in the group $GL(2, \mathbb{F_2})$. So why is the action the same as the symmetric group $S_3$?
Vectors in $\Bbb{F}_2^2$ are not the same as matrices in $\operatorname{GL}(2,\Bbb{F}_2)$. The set of non-zero vectors in $\Bbb{F}_2^2$ is $$\left\{\binom10,\binom01,\binom11\right\}.$$ Now check how each of the six elements of $\operatorname{GL}(2,\Bbb{F}_2)$ you listed acts on this set. For example $$\begin{pmatrix}0&1\\1&0\end{pmatrix}\qquad\text{ and }\qquad\begin{pmatrix}1&1\\0&1\end{pmatrix},$$ permute the first two elements and the last two elements listed above, respectively. What does that tell you about the group they generate?