Let $G$ be a Lie group and $H$ a closed subgroup. Then $G/H$ has a unique structure of a smooth manifold with canonical projection $p: G \to G/H$. If $\mathfrak g = T_e(G), \mathfrak h = T_e(H)$ are the Lie algebras of the groups.
1) How does the adjoint action of $H$ on $\mathfrak g$ defines an action $\psi$ of $H$ on $\mathfrak g/ \mathfrak h$?
2) How does $G$ act on the space of vector fields on $G/H$?
3) How does this lead to a linear isomorphism $T_{gH}(G/H) \to \mathfrak g/\mathfrak h$?
4) How do we now conclude that there is a bijection $\mathfrak X(G/H) \cong \{f \in C^\infty(G,\mathfrak g/\mathfrak h): f(gh) = \psi(h^{-1})(f(g)) \quad \forall g \in G, h \in H\}$?
My ideas: 'ad' 1) I guess I have to show that the subspace $\mathfrak h$ is $H$-invariant. If I can do this, the action descends to the quotient. So I have to show that for any $X \in \mathfrak h$ and any $h \in H$, we have $Ad(h)(X) \in \mathfrak h$. But $\mathfrak h = \{X \in \mathfrak g: \exp(tX) \in H \; \forall t\}$ and $\exp(t Ad(h)(X)) = h \exp(tX)h^{-1} \in H$, so this proves 1) (ok?)
'ad' 2) I am a bit lost here. If I have an action of $G$ on a manifold $M$ I get a map $\mathfrak g \to \mathfrak X(M), X \mapsto X_M$, where $X_M$ is the vector field defined by $X_M|_p = \frac{d}{dt}(exp(tX)\cdot p)|_{t=0}$. Does this help me here? I could let $G$ act on $M = G/H$ by left translation. How to proceed?
'ad' 3),4) First I have to understand 1) and 2) but I take any hints :-)
Your agrument for 1) is correct, alternatively, you can observe that the restriction of $Ad(h)$ to $\mathfrak h\subset\mathfrak g$ is the adjoint action of the Lie group $H$ on its Lie algebra.
2) is just a special case of the fact that a diffeomorphism of a manifold can be used to push forward vector fields. Explicitly, for $\xi\in\mathfrak X(G/H)$ and $g\in G$, you define $(g\cdot\xi)(\tilde gH):=T_{g^{-1}\tilde gH}\ell_g\cdot\xi(g^{-1}\tilde gH)$, where $\ell_g:G/H\to G/H$ is the left action of $g$. Otherwise put $g\cdot\xi=T\ell_g\circ\xi\circ\ell_{g^{-1}}$, which easily shows that this is indeed a right action.
3) is does not really depend on 2) but mainly on the fact that $p:G\to G/H$ is a surjective submersion. This implies that given $g\in G$, the map $T_gp:T_gG\to T_{gH}G/H$ is surjective. Now $X\mapsto L_X(g)$, where $L_X$ is the left invariant vector field generated by $X\in\mathfrak g$ is a linear isomorphism $\mathfrak g\to T_gG$, so $X\mapsto T_gp\cdot L_X(g)$ defines a surjection $\mathfrak g\to T_{gH}G/H$. Now the flow of $L_X$ throuh $g$ is given by $t\mapsto g\exp(tX)$, which easily shows that the kernel of this linear map is $\mathfrak h$. Hence it descends to a linear isomorphism $\phi_g:\mathfrak g/\mathfrak h\to T_{gH}G/H$.
4) Given a vector field $\xi\in\mathfrak X(G/H)$, the corresponding function $f:G\to \mathfrak g/\mathfrak h$ is defined by $f(g):=\phi_g^{-1}(\xi(gH))$. For $h\in H$, we get $ghH=gH$, so both $\phi_g$ and $\phi_{gh}$ are linear isomorphisms $\mathfrak g/\mathfrak h\to T_{gH}G/H$. From the construction in 3) one verifies that $\phi_{gh}=\phi_g\circ Ad(h)$, which implies that $f(gh)=Ad(h^{-1})(f(g))$, so $f$ is equivariant. Using local smooth sections of $p$, one can verify directly that $f$ is smooth. Conversely, given a smooth equivariant function $f: G\to \mathfrak g/\mathfrak h$, you observe that $\xi(gH):=\phi_g(f(g))$ is well defined by equivariancy and verify that it is a smooth vector field on $G/H$.
All this is a special case of the description of sections of a homogeneous vector bundle and, more generally, sections of an associated vector bundle to a principal fiber bundle.