I am trying to show the below statement which I very strongly feel should be "obviously correct", but I think I am missing the easy way to see this.
Let $G$ be a Lie group and $H$ a closed subgroup. By the closed subgroup theorem, $H$ is an embedded submanifold of $G$ (the same is true for each orbit $gH$ right?). Pick any $g\in G$ and let $\theta_g : G\to G$ be the diffeomorphism given by left-action of $g$.
We can start with $T_e H \subset T_e G$ and compute $d\theta_g (T_{e}H)$ which will be a subspace of $T_{g}G$. But will it be equal to $T_{g} (gH)$?
[I can only show that they are of the same dimension because $d\theta_g$ is an isomorphism.]
Let $v\in T_eH$, so $v$ can be represented by a path $\gamma:(-\epsilon,\epsilon)\to H$ with $\gamma(0)=e$. By definition, $d\theta_g(v)$ is represented by the path $\delta$, given by $$\delta(t)=g\gamma(t).$$Since the image of $\delta$ is contained in $gH$, we conclude$$d\theta_g(v)\in T_ggH,$$and so $$d\theta_g(T_eH)\subset T_ggH.$$The argument you gave, regarding the dimensions, should finish the proof.