If $H$ is a subgroup of $G$ and $G/H$ is the associated homogeneous space, then we have a fiber sequence $H \to G \to G/H \to BH \to BG$. In particular, $G/H$ is homotopy equivalent to the total space of a principal $G$-bundle over $BH$.
Question: Is there a simple but nondegenerate example where this principal $G$-bundle $G \to G/H \to BH$ is realized explicitly?
Ok, I'm not entirely sure this is what you are looking for, but you can't get any more explicit that this. I will provide a principal $G$-bundle $\pi:E\rightarrow BH$ with a very simple homotopy equivalence $E\simeq G/H$.
To this end let $E$ be the pullback of the cospan
$EG\times_G(G/H)\xrightarrow{p} BG\xleftarrow{\pi_G} EG\qquad(*)$
where the left-hand arrow is induced by projection onto the first factor and the right hand map belongs to the universal bundle for $G$.
First note that $EG\times_G(G/H)\cong (EG)/H\simeq BH$ where the first isomorphism is induced by multiplication and the second homotopy equivalence holds since $H\leq G$ acts freely on the contractible space $EG$. This homotopy equivalence can be made explicit by considering the $H$-equivariant inclusion $EH\xrightarrow{Ei}EG\hookrightarrow EG\times_G(G/H)$ ($H$ acts trivially on the right-most space), and from this it is clear that the homotopy equivalence is the induced map $BH\rightarrow EG\times_G(G/H)$. From this it is clear that the composition of this map with $p$ is homotopic to the map $Bi$ classifying the inclusion $i:H\leq G$.
Now the (strict) pullback $E$ carries a free right $G$-action induced from that on the universal $G$-bundle which, by construction, turns it into the principal $G$-bundle induced $Bi:BH\rightarrow BH$. Moreover since $EG$ is contractible and $\pi_G$ a fibration we see that the (strict) pullback of $(*)$ has the homotopy type of the homotopy fibre of $Bi$ (which we know to be $G/H$, although this will be made much more obvious). The end result is a principal $G$-bundle $\pi$ classified by $Bi$
$G\hookrightarrow E\xrightarrow{\pi}EG\times_G(G/H)\left(\simeq BH\xrightarrow{Bi} BG\right)$
Now what I believe will be disappointing for you is a classic statement, found in Proposition 14.1.3 of T. tom Dieck's Algebraic Topology (page 329-330):
The map $\theta:EG\times (G/H)\rightarrow E$ induced by $pr_1:EG\times (G/H)\rightarrow EG$ and the quotient $q:EG\times (G/H)\rightarrow EG\times_G (G/H)$ is a homeomorphism.
The inverse is constructed in the book. Putting everything together we get the following. The quotient is a principal $G$-bundle
$q:EG\times(G/H)\rightarrow EG\times_G(G/H)$
with total space homotopy equivalent to $G/H$ and base space homotopy equivalent to $BH$. With respect to these equivalences the bundle is classified by the map $Bi:BH\rightarrow BG$.