Consider the path $D_s(T)=A(s)B(t)A(s)^{-1}B(t)^{-1}$ in $G$ for some fixed value of s. Then the Lie bracket $[X,Y]$ can be related to the commutator of $A(s)B(t)A(s)^{-1}B(t)^{-1}$ of smooth paths $A(s)$ and $B(t)$ through $1\in G$.
The question asks how do we know $D'_s(t)\in T_1(G)$ and as $s$ varies, $D'_s(0)$ is a smooth path in $T_(G)$.
In this setting $A(s)$ and $B(t)$ are smooth paths in $G$ with $A(0) = B(0) = 1 =$ neutral element of $G$.
As $A$ and $B$ are smooth, as well as the group product $G \times G \to G, \, (g,h) \to g \cdot h$, the map $\psi: (s,t) \to D_s(t)$ is smooth, especially is $D_s(t)$ a smooth path for every fixed $s$. Plugging in $t=0$ gives $D_s(0) = A(s) 1 A(s)^{-1} 1 = 1$ and we conclude $D_s'(0) \in T_1 G$. The smootheness of $\psi$ imidiatly yields that $D_s'(0)$ is a smooth path in $T_1G \subset TG$.