Let $G$ be a locally compact second countable topological group and $H$ be a closed subgroup of $G$. Let $\pi: G\to G/H$ be the quotient map. For the Borel $\sigma$algebra $\mathcal A$ on $G/H$. Is it true that $\pi^{1}(\mathcal A)$ is the Borel subsets of $G$ consisting of right $H$ orbits?
Yes. $\DeclareMathOperator{\calB}{\mathcal{B}}$ Since the projection $\pi: G \to G/H$ is continuous, preimages of Borel sets are Borel, so we have that $\pi^{1}{(\mathcal{A})}$ is contained in the $\sigma$algebra $\calB_{H}$ of right $H$invariant Borelsets on $G$. To see the reverse inclusion $\calB_{H} \subset \pi^{1}(\mathcal{A})$, it suffices to show that $\pi(B) \in \mathcal{A}$ for every $B \in \mathcal{B}_{H}$ since for such $B$ we have $\pi^{1}(\pi(B)) = B$. _{It is not true that $\pi$ maps Borel sets to Borel sets. The image of a Borel set under $\pi$ will only be analytic in general. Historically, this remark can be seen as the starting point of descriptive set theory: Lebesgue famously assumed in an argument that the projection of a Borel set in $\mathbb{R}^{2}$ to one of its coordinates is Borel in $\mathbb{R}$. Sierpinski gave an example showing that this is wrong.} It is a basic but nontrivial result in descriptive set theory that there is a Borel transversal for the right $H$action on $G$, that is to say, there is a Borel set $T \subset G$ meeting every $H$coset exactly once $$ see Kechris, Classical descriptive set theory, Springer GTM vol. 156, Theorem 12.17. Observe that the restriction $\pi: T \to G/H$ is a Borel measurable bijection, hence it maps Borel subsets of $T$ to Borel sets in $G/H$ (see Kechris, Corollary 15.2). In other words, $\pi$ yields a Borel isomorphism $T \to G/H$. For every $B \in \calB_{H}$ we have $\pi(B \cap T) = \pi (B)$ and because $B \cap T$ is Borel we have $\pi(B) = \pi(B \cap T) \in \mathcal{A}$. 

