Fibered product of stacks comes from a Lie groupoid I am adding some context here. I am reading Introduction to Differentiable Stacks by Gregory Ginot.
In page no $7$, just before the remark $2.2$ he says the following.

One shall be careful that fibre product of differentiable stacks is not necessarily a differentiable stack  (though it is a stack over $\text{Diff}$).

So, one question  we can think of is, 

when does a fibre product of differentiable stacks a differentiable stack? 

This seems to be too difficult to answer. So, I look for a simpler question.

When does $B\mathcal{G}\times_{B\mathcal{H}} B\mathcal{G}$ a differentiable stack where the fibre product is coming from a given morphism of stacks $B\mathcal{G}\rightarrow B\mathcal{H}$.


Question in context of Lie groups is as follows. 
Suppose $G,H$ be Lie groups and $F:BG\rightarrow BH$ be a map of stacks.

When can we say that  the fibered product $BG\times_{BH}BG$ is a differentiable stack i.e., of the form $B\mathcal{K}$ for some Lie groupoid $\mathcal{K}$?

Assume that this  $F:BG\rightarrow BH$ is coming from a morphism of Lie groups $\theta:G\rightarrow H$. 
Then, we can ask following question.

When can we say that  the fibered product $BG\times_{BH}BG$ is a differentiable stack  coming from Lie group i.e., of the form $BK$ for some Lie group $K$?

Any comments are welcome.
 A: Pullbacks of stacks coming from Lie groupoids are not always equivalent to Lie groupoids.
Take $G=H=\mathbb{R}$.  Define $F(x)=0$ if $x\leq 0$ and $F(x)=exp(−1/x^2)$ if $x>0$.
The pullback is not equivalent to a Lie groupoid in this situation:
the set-theoretical pullback is $(−\infty,0]\times(−\infty,0]\cup \{(x,x)|x\in \mathbb{R}\}$,
which is clearly not a smooth manifold.
One can guarantee that the pullback is a Lie groupoid by imposing transversality
conditions on the maps involved.
That is, $A \times_C B$ is a Lie groupoid if $A_0 \rightarrow C_0 \leftarrow B_0$ is transversal
and $A_1 \rightarrow  C_1 \leftarrow B_1$ is transversal,
where subscripts $0$ and $1$ denotes objects and morphisms respectively.
(In fact, this transversality condition guarantees that the pullback
is also a homotopy pullback, which is almost always what one actually wants.)
A: Think of $BG$ and $BH$ as topological stacks, whereby one can calculate a topological groupoid presenting the stack $BG\times_{BH} BG$, namely the following: the object space is the space underlying $H$ and the morphism space is $G\times H \times G$. The source map $s\colon G\times H \times G \to H$ is the projection on the middle factor; the target map $t\colon G\times H \times G \to H$ is $(g_1,h,g_2) \mapsto \theta(g_1)^{-1}h\,\theta(g_2)$. (The reason you can do this is because the Lie groupoids presenting the stacks $BG$ and $BH$ only have a single object, so the situation is rather special. For more general Lie groupoids it is a little more fiddly, but not too different.)
For the stack $BG\times_{BH} BG$ to be equivalent to one of the form $BK$ for some topological group $K$, the topological groupoid I just described must be transitive: every object should be isomorphic to every other object. That is, for every pair $h_1,h_2\in H$ there should be elements $g_1,g_2\in G$ such that $h_2 = \theta(g_1)^{-1}h_1\theta(g_2)$. If $\theta$ is surjective, then you can take $g_1 = e_G$ and $g_2$ any lift of $h_1^{-1}h_2$, so this is a sufficient condition (in the topological case).
Now if we go back to general $\theta$, but take the special case $h_1 = e_H$, then we require for any $h\in H$ that $h= \theta(g_1^{-1}g_2)$, so in fact $\theta$ surjective is a necessary condition. To figure out which group $K$ it is such that $BG\times_{BH} BG \simeq BK$—it is a priori well-defined up to isomorphism of topological groups—we need to consider the pairs $g_1,g_2$ such that $h = \theta(g_1)^{-1}h\theta(g_2)$ for some chosen $h\in H$. We might as well take $h=e_H$, so that we want pairs $g_1,g_2 \in G$ such that $e_H = \theta(g_1^{-1}g_2)$, or in other words, such that $\theta(g_1) = \theta(g_2)$. The space of such pairs is just $G\times_H G$, which is a topological subgroup of $G\times G$, and so $K=G\times_H G$.
(In fact $G\times \ker \theta \to G\times_H G$, $(g,k) \mapsto (g,gk)$ is a homeomorphism, but only a topological group isomorphism if $\ker \theta \lt G$ is central subgroup.)

Now if we want to do this in Lie groups, then everything works, except that we need $K = G\times_H G$ to be a sub-Lie-group of $G\times G$, and this is so if $\theta$ is a submersion. But a surjective map of (finite-dimensional) Lie groups is automatically a submersion. Thus $G\to H$ is a surjective submersion, and hence a locally trivial bundle (this follows from using charts derived from the exponential map and the surjective map of the associated Lie algebras).
If we don't care about $BG\times_{BH} BG \simeq BK$ for some $K$, then $\theta$ being a submersion should be enough to make $G\times H \times G \rightrightarrows H$ a Lie groupoid (the only hard part is to show that $(g_1,h,g_2) \mapsto \theta(g_1)^{-1}h\,\theta(g_2)$ is a submersion).
Even in the special case analysed above, we don't know that $K$ is a central subgroup, or that $G\twoheadrightarrow H$ is a central extension, but that's not necessary for your question. So we find ourselves in the situation Dmitri gave in greater generality: $(G\rightrightarrows \ast) \to (H \rightrightarrows \ast)$ is a submersion on arrows and object components.

Added  There was a small unimportant lie in what I wrote: it is not sufficient in the topological case for $\theta$ to be just surjective, to have an equivalence of stacks $BG \times_{BH} BG \simeq BK$. Namely, it is not true that the groupoid that I described being transitive is enough. One must have a certain map to have local sections, and this boils down to requiring that $\theta$ have local sections. In the smooth case we ask that this map is a surjective submersion, but this follows from $\theta$ just being surjective, as this already implies it is a submersion.
