One way to see Cramer's rule is that it simply makes use of a (very inefficient) way of calculating $A^{-1}$, specifically \begin{equation}A^{-1}=\frac{1}{\text{det}(A)}\text{Adj}(A),\end{equation} where Adj($A$) is known nowadays as the adjugate of $A$ (transpose of matrix of cofactors of $A$). If you substitute this equation into $X=A^{-1}B$ you basically get Cramer's rule, since for every component $x_i$ of $X$ you get the Laplace expansion by the ith column of $1/$det$(A)\cdot$det$(A_{C_i\leftrightarrow B})$ (you can check this...).
So then to interpret this geometrically it depends on how far you want to go. If you just see it as a way of calculating $A^{-1}B$ then it simply means that a vector $X$ which is transformed by $A$ to give the vector $B$, if this transformation is invertible then you can calculate $X$ by applying the inverse transformation to $B$. If you now want to go further you have to try and explain geometrically why \begin{equation}A^{-1}=\frac{1}{\text{det}(A)}\text{Adj}(A).\end{equation} So in terms of a geometrical interpretation, we know $A$ is a transformation, and det($A$) is the area of the parallelogram formed by the component vectors in $A$. So what can be interesting is to explore the role of the adjugate and how it determines the formulae for det$(A)$...if you want to...I'll leave this for you - in 2D it's simple, since the cofactors of each entry is just another entry (with a sign change where needed).
