Let $G$ be a compact Lie group and $H$ be a closed subgroup. The inclusion $H \rightarrow G$ induces a homotopy fibration $G/H \rightarrow BH \rightarrow BG$. In particular, this must hold if $G$ and $H$ are finite groups. For example, if $G = \mathbb{Z}/2 \times \mathbb{Z}/2$ and $H = \mathbb{Z}/2$, then there is a fibration $\mathbb{Z}/2 \rightarrow \mathbb{R}P^\infty \rightarrow \mathbb{R}P^\infty \times \mathbb{R}P^\infty$
does this mean that there is a surjective map of $\mathbb{R}P^\infty \rightarrow \mathbb{R}P^\infty \times \mathbb{R}P^\infty$, and is this map a double covering? I hardly think that the last statements are true, so there should be something wrong along the way that I am missing.
The sequence $G/H\to BH\to BG$ is a homotopy fiber sequence. That means the map $BH\to BG$ is not necessarily literally a fibration with fiber $G/H$; rather, this is true up to homotopy equivalence. That is, there is a fibration $E\to BG$ with fiber $F\to E$ and homotopy equivalences $BH\to E$ and $G/H\to F$ making the obvious diagram commute.
In your example, you can see this quite explicitly. Supposing $H=\mathbb{Z}/2$ is the first coordinate of $G=\mathbb{Z}/2\times\mathbb{Z}/2$, then $BG=\mathbb{R}P^\infty\times\mathbb{R}P^\infty$ has a double cover $\mathbb{R}P^\infty\times S^\infty\to\mathbb{R}P^\infty\times\mathbb{R}P^\infty$. Since $S^\infty$ is contractible, this double cover is homotopy equivalent to $BH=\mathbb{R}P^\infty$ in a natural way.
More generally, if $G$ is a finite group, then $BG$ has a covering space associated to the subgroup $H\subseteq\pi_1(BG)\cong G$. This covering space may not be literally the same as whatever specific definition of $BH$ you are using, but it will be homotopy equivalent, since it is a space whose only nontrivial homotopy group is $\pi_1$, which is $H$.