The tangent bundle of G/H as an associated bundle

Define $\mathrm{\Theta }([g,X+\mathfrak{h}])=d(L_g^{G\mathrm{/}H}{)}_{eH}(\overline{d{q}_e}(X+\mathfrak{h}))\mathrm{.}$ This vector lies over $gH$. The formula is representative-independent. Indeed, $[gh,v]=[g,{\mathrm{Ad}}_{h}v]$ in the associated bundle, while [F1] gives $d(L_{gh}{)}_{eH}\overline{d{q}_e}(v)=d(L_g{)}_{eH}d(L_h{)}_{eH}\overline{d{q}_e}(v)=d(L_g{)}_{eH}\overline{d{q}_e}({\mathrm{Ad}}_{h}v)\mathrm{.}$

On the fibre over $gH$, $\mathrm{\Theta }$ is the composite of the linear isomorphisms $\overline{d{q}_e}$ and $d(L_g{)}_{eH}$, so it is a fibrewise-linear bijection. In a principal trivialization with smooth section $s:U\to G$, its coordinate expression is $(x,v)⟼d(L_{s(x)}{)}_{eH}\overline{d{q}_e}(v)\mathrm{.}$ This is smooth. Its inverse applies $d(L_{s(x{)}^{-1}}{)}_x$ and then ${\overline{d{q}_e}}^{-1}$, so it is smooth as well.

Hence $\mathrm{\Theta }$ is a smooth vector-bundle isomorphism over $G\mathrm{/}H$. If $H=G$, both sides are the zero bundle over a point; if $H=\{e\}$, this is the standard left trivialization $G\times \mathfrak{g}\cong TG$. Normality of $H$ is not needed; it is precisely the isotropy action, not an action assumed trivial, that makes step 1.1 work. Countable choice is inherited through [A1] and [F1].