What is the geometrical intuition of defining the Lie bracket on $\mathfrak g$ via the adjoint representation?

Question

Let $G$ be a Lie group, and its Lie algebra $\mathfrak g\equiv\mathrm{Lie}(G)\equiv T_e G$ be defined as its tangent space at the identity, where tangent vectors are defined as equivalence classes of curves.

(Lie bracket via left-invariant vector fields) One way to define the Lie bracket in $\mathfrak g$ is via the left-invariant vector fields: if $X,Y\in\Gamma(TM)$ are left-invariant, and we denote with $\nabla_X,\nabla_Y\in\mathrm{Der}(C^\infty(M))$ the corresponding derivations, we can define $[X,Y]\in\Gamma(TM)$ via $$\nabla_{[X,Y]} = [\nabla_X,\nabla_Y]\equiv \nabla_X\circ\nabla_Y-\nabla_Y\circ\nabla_X.$$ Therefore, given $\gamma'(0),\eta'(0)\in T_e G$, if $X_\gamma$ is the left-invariant vector field such that $X(e)=(e,\gamma'(0))$, then we can define $$[\gamma'(0),\eta'(0)] \equiv \eta_2([X_\gamma,Y_\eta](e)),$$ where $\eta_2:TG\to T_e G$ projects onto the second component: $\eta_2((p,v))\equiv v$ for all $p\in M, v\in T_p M$ (notation is from Terry Tao's notes).

(Lie bracket via adjoint representation) As also discussed in this other answer, one can also define $[\gamma'(0),\eta'(0)]$ without explicitly passing through left-invariant vector fields, by leveraging the adjoint representation of the group. More precisely, we can build this as follows

1. Consider the map $\Psi_g:G\to G: g\mapsto hgh^{-1}$. Then its differential at the identity reads $$\mathrm{Ad}(g) \equiv \mathrm d(\Psi_g)_e: \mathfrak g\to \mathfrak g.$$
2. Observe that $\mathrm{Ad}:G\ni g\mapsto \mathrm{Ad}(g)\in \mathrm{GL}(\mathfrak g)$. Explicitly, we thus have $$\mathrm{Ad}(g): \mathfrak g\ni \gamma'(0)\mapsto (g\gamma g^{-1})'(0) \equiv \partial_t|_{0} (g\gamma(t) g^{-1})\in \mathfrak g.$$ Note also that $(\mathrm{Ad}(e))(\gamma'(0)) = \gamma'(0)$, that is, $\mathrm{Ad}(e)=I\equiv \mathrm{Id}_{\mathfrak g}$.
3. We now define $\mathrm{ad}:\mathfrak g\to\mathrm{End}(\mathfrak g)$ as $\mathrm{ad} \equiv \mathrm d(\mathrm{Ad})_e : \mathfrak g\to T_{I}\mathrm{GL}(\mathfrak g)$, explicitly, its action reads $$\mathrm{ad}(\gamma'(0)) \equiv \partial_s|_{0}\, (\mathrm{Ad}(\gamma(s))), \\ [\gamma'(0),\eta'(0)] \equiv (\mathrm{ad}(\gamma'(0)))(\eta'(0)) = \partial_s|_0 \partial_t|_0 \, (\gamma(t)\eta(s)\gamma^{-1}(t)).\tag X$$

(The question) My question is specifically about the second expression in (X). Assuming I'm not mistaken in this, we can define directly the commutator between two tangent vectors, understood here as (equivalence classes of) curves in the manifold, as a double derivative of the product $\gamma(t)\eta(s)\gamma^{-1}(t)$.

Is there a more direct way to see why this particular definition should work? If I'm reading this correctly, this is saying that the commutator of (the tangent vectors corresponding to) the curves $\gamma$ and $\eta$, corresponds to the curve $$s\mapsto \partial_t|_0 \, (\gamma(t) \eta(s) \gamma^{-1}(t)). \tag Y$$ Is there a more direct/intuitive/geometric way to understand the significance of this particular curve?