Let $G$ be a Lie group and $H$ a closed subgroup with Lie algebras $\mathfrak g, \mathfrak h$. Then the canonical projection $p: G \to G/H$ is a submersion. Fix a $g \in G$. We have a linear isomorphism $\omega(g): T_g G \to \mathfrak g$ given by the Maurer-Cartan form $\omega$, i.e. $\omega(g)(X) = T_g \lambda_{g^{-1}}$ ($\lambda_g$ left multiplication with $g$). Further we have the projection $\pi: \mathfrak g \to \mathfrak g/\mathfrak h$.
I want to show that we have a linear isomorphism $\varphi_g: T_{gH}(G/H)\to \mathfrak g/\mathfrak h$. To define it, I want to define $\varphi_g$ such that $\varphi_g \circ T_g p = \pi \circ \omega(g)$. For this to be well-defined (and injective) I should make clear to myself why $\ker T_g p = \omega^{-1}(\mathfrak h)$. Did I get this right? ($\omega^{-1}(g) = T_e \lambda_g$, right?)
I've read somewhere this line: $\ker T_e p = T_e(p^{-1}(eH)) = T_e H = \mathfrak h$, i.e. the kernel of $T_e p$ is the tangent space at $e$ of the fiber $p^{-1}(eH)$. Why is that? And how to continue the calculation $\ker T_g p = T_g(gH)$? That is, why is $T_g(gH) = T_e \lambda_g (\mathfrak h) (=\omega^{-1}(\mathfrak h))?$
There is a general result you should be familiar with. Let $f \colon M \rightarrow N$ be a smooth map and let $q \in N$ be a regular value of $f$. Then $f^{-1}(q)$ is a smooth submanifold of $M$ and if $p \in f^{-1}(q)$ then $T_p(f^{-1}(q)) = \ker df_p$. To see this, note that if $\gamma \colon I \rightarrow f^{-1}(q)$ is a smooth curve with $\gamma(0) = p$ then
$$ df_p(\dot{\gamma}(0)) = \frac{d}{dt} f(\gamma(t))|_{t = 0} = \frac{d}{dt} q|_{t = 0} = 0. $$
This shows that $T_p(f^{-1}(q)) \subseteq \ker df_p$ and since the dimensions of the two vector spaces are equal, we must have equality.
In your case, $p$ is a submersion so any value is a regular value and you have $$T_g(gH) = T_g(p^{-1}(gH)) = \ker dp_{g}. $$