Say one has 2 vectors in 2D $v_1, v_2$. One can construct the bivector $b_1 = v_1 \wedge v_2$. Say that there is a linear map $L$ that relates 2 other vectors as follows:
$L(v_1) = u_1, L(v_2) = u_2$.
$L$ acting on vectors this way implies the existence of a map acting on the bivectors themselves as:
$M(b_1)=L(v_1)\wedge L(v_2) = u_1\wedge u_2 = b_2$
If one has a coordinate representation of $v, u, L$, is there a coordinate represnetation of $M$? Is it different from $L$?
Considering $\Lambda^2\mathbb{F}^2$ is 1-dimensional, you don't need coordinates; $M$ is a $1\times1$ matrix (as opposed to $L$, which must be represented by a $2\times2$ matrix). Indeed, $M=[\det L]$.
In general, $\dim\Lambda^k\mathbb{F}^n=\binom{n}{k}$. It is a standard exercise to show
$$\{e_{i_1}\wedge\cdots e_{i_k}\mid 1\le i_1<\cdots<i_k\le n\}$$
is a basis (which may be lexicographically ordered), which is in bijection with $k$-subsets of $\{1,\cdots,n\}$. In particular, we always have $\dim\Lambda^n\mathbb{F}^n=1$. If $A$ is any linear operator acting on $\mathbb{F}^n$ then it induces an action on multivectors, $$A(v_1\wedge\cdots\wedge v_k)=(Av_1)\wedge\cdots\wedge(Av_k).$$
In this setting, $A$ acting on the top exterior power $\Lambda^n\mathbb{F}^n$ is just a scalar transformation, namely acting as multiplication-by-$\det A$.
If you represent $v_1,v_2,u_1,u_2$ with respect to some chosen coordinates and write $U=(u_1~u_2)$, $V=(v_1~v_2)$ as matrices, then we may explicitly compute $L=UV^{-1}$.