I'm a bit confused by the double role which sheaves play in the theory of stacks.


On the one hand, sheaves on a site are the obvious generalization of a sheaf on a topological space. On the other hand a sheaf on a site is (or better its associated category fibered in sets is) a very particular stack itself, so a generalization of a space. This is not completely confusing: more or less it amounts (I believe) to identifying a space X with the sheaf of continuos functions with values in X.


But now my question is the following. An equivalent condition for a fibered category to be a prestack is that for any two objects (over the same base object), the associated functor of arrows should be a sheaf. In particular this is true for a stack, so for any stack and any two objects in it we have a sheaf, and so a stack (over a comma category).


What is the meaning of this geometrically?


For instance take the stack $\mathcal{M}\_{g,n}$.


Giving two objects in the stack (over the same base object) means giving two families $X$ and $Y$ of stable pointed curves over the same scheme $S$, and the associated functor of arrows maps every other scheme $f \colon T \rightarrow S$ to the set of morphism between $f^\* X$ and $f^\* Y$. How should I think of the associated stack as a space?


To avoid misunderstandings I give the defition of the functor of arrows. Let $\mathcal{F}$ be a fibered category over $\mathcal{C}$. Take $U \in \mathcal{C}$ and $\xi, \eta \in \mathcal{F}(U)$. Then there is a functor $F \colon \mathcal{C}/U \rightarrow Set$ defined as follows. For a map $f \colon T \rightarrow U$ we put $F(f) = Hom(f^\* \xi, f^\* \eta)$. The action on arrows requires some diagrams to be described, but it's really the only possible one.



Stacks and sheaves
Let me see if I understand your example correctly: you are fixing $X$ and $Y$, families
of curves over $S$, and now you are considering the functor which maps an $S$-scheme $T$
to the set of $T$-isomorphisms $f^\*X \to f^\*Y$ (where $f$ is the map from $T$ to $S$).


If I have things straight, then this functor shouldn't be so bad to think about, because it is actually representable, by an Isom scheme. In other words, there is an $S$-scheme
$Isom\_S(X,Y)$ whose $T$-valued points, for any $f:T \to S$, are precisely the $T$-isomorphisms
from $f^\*X$ to $f^\*Y$. (One can construct the Isom scheme by looking inside a 
certain well-chosen Hilbert scheme.)


One way to think about this geometrically is as follows: one can imagine that two
curves over $k$ (a field) are isomorphic precisely when certain invariants coincide
(e.g. for elliptic curves, the $j$-invariant). (Of course this is a simplification,
and the whole point of the theory of moduli spaces/schemes/stacks is to make it precise,
but it is a helpful intuition.) Now if we have a family $X$ over $S$, these invariants
vary over $S$ to give a collections of functions on $S$ (e.g. a function $j$ in the
genus $1$ case), and similarly with $Y$. Now $X$ and $Y$ will have isomorphic
fibres precisely at those points where the invariants coincide, so if we look
at the subscheme $Z$ of $S$ defined by the coincidence of the invariants,
we expect that $f^\*X$ and $f^\*Y$ will be isomorphic precisely if the map $f$
factors through $Z$. Thus $Z$ is a rough approximation to the Isom scheme.


It is not precisely the Isom scheme, because curves sometimes have non-trivial
automorphisms, and so even if we know that $X\_s$ and $Y\_s$ are isomorphic for
some $s \in S$, they may be isomorphic in more than one way. So actually the
Isom scheme will be some kind of (possibly ramified) finite cover of $Z$.


Of course, if one pursues this line of intuition much more seriously, one will
recover the notions of moduli stack, coarse moduli space, and so on.


Added: The following additional remark might help: 


The families $X$ and $Y$ over $S$ correspond to a map $\phi:S \to {\mathcal M}\_g
\times {\mathcal M}\_g$. The stack which maps a $T$-scheme to $Isom\_T(f^\*X,
f^\*Y)$ can then seen to be the fibre product of the map $\phi$ and the diagonal
$\Delta:{\mathcal M}\_g \to {\mathcal M}\_g \times {\mathcal M}\_g$.


In the particular case of ${\mathcal M}\_g$ the fact that this fibre product is representable is part of the condition that ${\mathcal M}\_g$ be an algebraic stack.


But in general, the construction you describe is the construction of a fibre product
with the diagonal. This might help with the geometric picture, and make the relationship to Mike's answer clearer. (For the latter:note that the path space into $X$ has a natural
projection to $X\times X$ (take the two endpoints), and the loop space is the fibre product
of the path space with the diagonal $X\to X\times X$.)


