I think there's some confusion in the question. For example by "Levi-Civita connection" you must really mean some kind of Laplacian. Anyway, your end result about the cohomology of $G/H$ is essentially correct. To help clarify the situation, let's bring in some relative Lie algebra cohomology.
The first basic fact is that a Lie group $G$ acts naturally on the space $\Omega^\ast(G/H)$ of differential forms on $G/H$, where $H$ is a closed subgroup of $G$. If $G$ and $H$ are connected, then evaluation at the identity gives an isomorphism from the space $\Omega^\ast(G/H)^G$ of $G$-invariant forms onto the space $C^\ast(\mathfrak g, \mathfrak h)$ of relative Lie algebra cohomology cocycles. This isomorphism descends to an isomorphism between the cohomology of the complex of $G$-invariant forms on $G/H$ and relative Lie algebra cohomology $H^\ast(\mathfrak g, \mathfrak h)$.
Now assume that $G$ is reductive and $H$ maximal compact in $G$ with corresponding Cartan decomposition $\mathfrak g = \mathfrak h \oplus \mathfrak p$. Then there are natural isomorphisms
$$ C^q(\mathfrak g, \mathfrak h) = \mathrm{Hom}_{\mathfrak h} (\wedge^q \mathfrak p, \mathbb R) = (\wedge^q \mathfrak p)^{\mathfrak h}, $$
where $\mathfrak h$ is acting via the adjoint representation. If you want, this can be rephrased in terms of the action of $H$, and moreover it's possible to introduce a bilinear form so that the above isomorphisms yield
$$ C^q(\mathfrak g, \mathfrak h) = (\wedge^q \mathfrak h^\perp)^H. $$
The next key result is that the differential of this complex is zero, whence
$$ H^q(\mathfrak g, \mathfrak h) = (\wedge^q \mathfrak h^\perp)^H. $$
In some precise sense this is a manifestation of a theorem of E. Cartan which states that an invariant form on $G/K$ is automatically closed (and harmonic). Indeed, we can combine this last isomorphism with the isomorphisms mentioned in the second paragraph to conclude that
$$ \Omega^q(G/H)^G = (\wedge^q \mathfrak h^\perp)^H $$
and both are isomorphic to the cohomology of the complex of $G$-invariant forms on $G/H$.
In general we can't deduce from this that the cohomology of $G/H$ is isomorphic to $(\wedge^\ast \mathfrak h^\perp)^H$, because not every closed form will necessarily have an invariant form in its cohomology class. However, if $G$ is compact, then this is not a problem, because we can use the usual averaging trick.
Anyway, the point of introducing Lie algebra cohomology into the picture is that it makes things very computable. It will also be an important key phrase for any literature search.
