In what follows all the groups will be discrete, not necessarly finite.
Let $f:G\to H$ be a morphism of groups and $H'\to H$ be the inclusion of a subgroup. It seems to me (but correct me if I am wrong) that the pullback $BG\times_{BH}BH'$ is $B(f^{-1}H')$.
If this is true, is there a more general statement that tells you when a commutative square of classifying space is a pullback diagram?
On the other hand, if we define $BG$ by the usual simplicial construction, then the functor $B$ does indeed preserve pullbacks. Indeed, if we start with a pullback square $(G,H,J,K)$ then in the corresponding square of simplicial sets, the square of $n$-simplices is just $(G^n,H^n,J^n,K^n)$, which is again a pullback, so we have a pullback square of simplicial sets. It is also standard that the geometric realisation functor from simplicial sets to spaces preserves finite limits, so $(BG,BH,BJ,BK)$ is a pullback of spaces. If the map $H\to K$ is surjective then the map $BH\to BK$ is a fibration and we can conclude that the square $(BG,BH,BJ,BK)$ is also a homotopy pullback. The same applies if the map $J\to K$ is surjective. However, as the other answer explains, this story is not homotopy invariant. It is more usual to consider $BG$ as the name of a homotopy type, and with different models for the relevant homotopy types, the square $(BG,BH,BJ,BK)$ need not be either a pullback or a homotopy pullback.
This is not true. Note that $BG$ is only well-defined up to homotopy equivalence, so the only question that makes sense is when the square $$\begin{array}{ccc} B\big(G \underset H\times H'\big) & \to & BH' \\ \downarrow & & \downarrow \\ BG & \to & BH \end{array}$$ is a homotopy pullback square. For this, you could either work in topological spaces or in Kan complexes (i.e. $\infty$-groupoids). I find it easiest to take the latter point of view, since it is closer to group theory.
The homotopy pullback of a span $X \stackrel p\to Z \stackrel q\leftarrow Y$ can be computed as the space of triples $(x,y,\phi)$ of a point $x \in X$, a point $y \in Y$, and an isomorphism $\phi \colon p(x) \stackrel\sim\to q(y)$ (in the topological setting, you would take $\phi$ to be a path from $p(x)$ to $q(y)$). In the case of groups, there is no choice for the points, so the only choice is $\phi$. The morphisms $(*,*,\phi) \to (*,*,\psi)$ are given by pairs $(g,h') \in G \times H'$ such that $q(h') \circ \phi = \psi \circ p(g)$ (topologically, this would be a homotopy between the paths $p(x) \stackrel\phi\to q(y) \stackrel{q(h')}\to q(y')$ and $p(x) \stackrel{p(g)}\to p(x') \stackrel\psi\to q(y')$). Higher morphisms are determined by their 1-skeleta, so we again get a 1-truncated space, i.e. the nerve of a groupoid.
So what we get in the end is $H/(G \times H')$, where $(g,h') \in G \times H'$ acts by $h \mapsto q(h')gp(g)^{-1}$; the quotient being interpreted in the homotopical sense (as in $BG = */G$). For instance, we see that it is not always connected, so it won't be of the form $BG'$ for any group $G'$. This already happens when $H' = G = 1$ and $H \neq 1$: then the homotopy fibre product is just $H$ viewed as a discrete set.
In general, if $H' = 1$, then the fibre product will be the pullback of the universal cover of $BH$ to $BG$, which is connected if and only if $G \to H$ is surjective. The surjective case is also exactly when the map $N(G) \to N(H)$ (with the groups viewed as one-object categories) is a Kan fibration, so you can indeed compute the homotopy pullback as the ordinary pullback.
