Conjugation has a geometric meaning: $aba^{-1}$ is the transform $b$ under coordination change $a$. For example in the symmetry of tetrahedron, let $a$ be the anti-clockwise $1/3$ round rotation around the altitude through vertex $A$, $b$ be the reflection by the plane through edge $AB$, the $aba^{-1}$ is the reflection of the plane through edge $AC$, which is just $b$ after relabeling vertices $A$ $B$$C$ $D$ by $a$.
Is there a similar geometric meaning of the commutator, $aba^{-1}b^{-1}$?
This is arguably an algebraic fact more than a geometric one, but for any group $G$ and $a, b \in G$ rearranging gives that $$a b a^{-1} b^{-1} = 1_G \qquad \textrm{iff} \qquad a b = b a ,$$ so the commutator $a b a^{-1} b^{-1} \in [G, G]$ of $a, b$ is a measure of the failure of those two elements (or geometrically, symmetries) to commute and, if the elements don't commute, roughly speaking a "direction" of that failure.
An instructive example is the following: Consider the group $SO(3)$ of oriented rotations of $\Bbb R^3$. Small anticlockwise rotation about the $x$- and $y$-axes are respectively given to first order by $$\pmatrix{1\\&1&-\epsilon\\&\epsilon&1} , \pmatrix{1&&\epsilon\\&1&\\-\epsilon&&1} ,$$ and forming their commutator gives to second order that $$\pmatrix{1&\epsilon^2\\-\epsilon^2&1\\&&1} ,$$ which is a small clockwise rotation about the $z$-axis. (Instead forming the commutator in the reverse order gives a small anticlockwise rotation about the $z$-axis instead.)
For Lie groups of positive dimension like $SO(3, \Bbb R)$, a convenient algebraic way to formalize the idea of working with "small" symmetries (in our example, symmetries that don't rotate the sphere too much) is to pass to the level of infinitesimal symmetries. We can identify these symmetries with the tangent space $T_{1_G} G$ of $G$ at the identity $1_G$. In the case of $SO(3, \Bbb R)$, we can identify $T_I SO(3, \Bbb R)$ with the space of $3 \times 3$ antisymmetric real matrices, and linearization of the group multiplication law (more precisely, of the conjugation map) gives a Lie bracket on $T_I SO(3, \Bbb R)$, a natural, bilinear, antisymmetric map that in this case coincides with the usual commutator $$(A, B) \mapsto [A, B] := AB - BA$$ of matrices. The vector space $T_I SO(3, \Bbb R)$ together with the Lie bracket operation defines the Lie algebra $\mathfrak{so}(3, \Bbb R)$ of $SO(3, \Bbb R)$.