Let $G$ be a locally compact Hausdorff group, $H$ a closed normal subgroup. To simplify matters we assume that the underlying topological space of $G$ has a countable base. Suppose a left Harr measure $\mu$ is explicitly given on $G$.
Can we explicitly construct a left Haar measure on $G/H$ using $\mu$?
The motivation is as follows. When $H$ is compact or discrete, the answer seems to be yes. Suppose $H$ is compact. Let $K$ be a compact subset of $G/H$. Then $\psi^{-1}(K)$ is compact where $\psi$ is the canonical homomorphism $G \rightarrow G/H$. So we can define $\nu(K) = \mu(\psi^{-1}(K))$. Then I think $\nu$ defines a left Haar measure on $G/H$. When $H$ is discrete, $\psi$ is a local homeomorphism. So we can define locally a Borel measure on $G/H$. Then I think we can get it globally.
I have no idea about the answer when $H$ is not compact or discrete.
I think what you wrote is wrong even if you assume $H$ is discrete. Take $G=\mathbb{R}$, $H=\mathbb{Z}$.
So I think in general the answer to your question is no. But if $H$ is a lattice then you have a fundamental domain $\Omega$ in $G$ such that the measure of $A\subseteq G/H$ is $\pi^{-1}(A)\cap \Omega$, where $\pi:G\to G/H$ is the projection.