If you ever wanted to construct the tangent bundle of a differentiable stack, it's relatively simple:
First, if $\mathbf{X}$ is a stack coming from a Lie groupoid $\mathcal{G}$, you could just say $\mathbf{TX}$ is the stack associated to the tangent groupoid $\mathcal{TG}$.
A more formal way is the following:
Consider the composite
$\text{Mfd} \stackrel{T}{\rightarrow} \text{VectorBundles} \stackrel{\text{forget}}{\rightarrow} \text{Mfd} \stackrel{\text{yoneda}}{\rightarrow} \text{St}(\text{Mfd}).$
by restricting its weak left Kan extension to stacks we get a 2-functor (by abuse of notation)
$\text{St}(\text{Mfd}) \stackrel{T}{\rightarrow} \text{St}(\text{Mfd}).$
Now, suppose $\mathbf{X}$ is a stack coming from a Lie groupoid $\mathcal{G}$. Then $\mathbf{X}$ is the weak colimit of the composition
$\Delta^{\rm op} \stackrel{N(\mathcal{G})}{\rightarrow}Mfd\stackrel{\text{yoneda}}{\rightarrow}\text{St}(\text{Mfd})$
(in fact even of its 2-truncation.)
Since $T$ is weak-colimit preserving, it follows that $T \mathbf{X}$ is the weak colimit of
$\Delta^{\rm op} \stackrel{N(\mathcal{TG})}{\rightarrow}\text{Mfd}\stackrel{\text{yoneda}}{\rightarrow}\text{St}(\text{Mfd})$
which is in turn just the stack associated to the tangent groupoid $\mathcal{TG}$. So both definitions agree.
Now suppose you wanted to define the cotangent stack. The first line of attack, done naively, seems to fail. The cotangent functor $T^*$ is contravariant so it doesn't send groupoid objects to groupoid objects. However, less naively, in the literature (for instance here: http://arxiv.org/PS_cache/arxiv/pdf/0905/0905.4318v2.pdf), one can define the cotangent groupoid of $\mathcal{G}$ to be a certain symplectic groupoid $\mathcal{T^{*}G}_1 \rightrightarrows \text{Lie}(\mathcal{G})^{*}$, where $\text{Lie}(\mathcal{G})^{*}$ is the dual Lie algebroid of $\mathcal{G}$. (Is this invariant under Morita equivalence?)
The second way would seem to be:
Consider the opposite of the composite
$\text{Mfd}^{\rm op} \stackrel{T^*}{\rightarrow} \text{VectorBundles} \stackrel{\text{forget}}{\rightarrow} \text{Mfd} \stackrel{\text{yoneda}}{\rightarrow} \text{St}(\text{Mfd}),$
$\text{Mfd} \to \text{St}(\text{Mfd})^{\rm op}$
and take its weak left Kan extension. (So $\mathbf{T^{*}X}$ is the weak limit of $T^{*} M$ over all $M \to \mathbf{X}$.)
The opposite of this 2-functor goes $\text{St}(\text{Mfd})^{\rm op} \to \text{St}(\text{Mfd})$.
How do these two notions of cotangent stack relate to each other? Is either reasonable? Is there a better notion than either of these?
I would like, for instance, if you have a Riemannian metric on an orbifold, to get an equivalence between its tangent stack and its cotangent stack.
Along these lines, the underlying space of the cotangent bundle of an orbifold is an orbifold (or is it a manifold, like its frame bundle?), so it should be represented by a orbifold groupoid. Is this the same as the cotangent groupoid?
Keep in mind, the correct definition of tangent stack should produce the correct notion of differentiable forms on a stack, and the cotangent stack should somehow have a symplectic structure.
Best Answer
This is a question near and dear to my heart, since it is actually what my thesis will be about. Let me denote the Lie algebroid of a groupoid $G$ with base $M$ by $A$.
If you're viewing the stack presented by $G$ as a smooth model for the quotient space $M/G$, then the first proposal doesn't really seem to give the right answer. Namely, the Lie algebroid structure on $A$ induces a Poisson structure on $A^*$, and the quotient space of the groupoid $T^*G$ you mention is just the space of symplectic leaves for this Poisson structure --- this space doesn't give the cotangent bundle of $M/G$ and doesn't have a natural symplectic structure of its own. (Of course, these quotients aren't necessarily smooth, but even if they are you don't get the right answer.) Indeed, if you start with a trivial groupoid $M$, the space you get is just $M$.
Probably the underlying question you first have to answer is: in what sense is $T^*$ a functor? As Andre says, if all you care about is etale maps there's no problem. In general, you still get something interesting if you change the target category for the $T^*$ functor to the so-called symplectic category, where objects are symplectic manifolds, and a morphism $M \to N$ is a lagrangian submanifold of $\overline{M} \times N$, where the bar denotes changing the sign of the symplectic structure. (The usual caveat applies: this is not really a category since compositions are not always well-defined, but we'll just ignore this for now.) Composition is just the usual composition of relations.
Now to a smooth map $f: X \to Y$ you can assign the lagrangian submanifold
$T^*f := \{ (x, df_x^*\xi, f(x), \xi) \} \subseteq \overline{T^*X} \times T^*Y$,
where $df$ is the tangent map. This then gives a nice functor from the category of smooth manifolds to the symplectic category. In particular, if $f: X \to Y$ is a diffeomorphism (or even just etale), then $T^*f$ is in fact the graph of an actual map.
The upshot is that if you start with a groupoid, you can apply this functor to everything in sight to get a "groupoid" in the symplectic category. (This idea is due to Alan Weinstein. I'm calling this a "groupoid" since it is not clear in what sense this object is actually a groupoid. In particular, the symplectic category doesn't have fiber products, so the notion of a groupoid object isn't defined.) If your groupoid is etale, this recovers Andre's suggestion. So, you can instead try to formulate the notion of the cotangent stack of a stack in terms of this object. This post has gone on long enough, so I won't go into details here, but numerous examples suggest that this is indeed the right thing (or at least "a" right thing) to consider. In particular, from this point of view the "symplecticness" of cotangent stacks comes from the fact that they live in the symplectic category (this can be made more precise).
I should also say that this construction is somehow related to the groupoid structure on $T^*G$ you mentioned, but it contains more information that just that.
Edit: Here's one example illustrating some of the above. Say $G$ is a group acting smoothly on $M$. Then the action lifts to a Hamiltonian action on $T^*M$ with momentum map $\mu$. When the action is free, the resulting symplectic quotient $\mu^{-1}(0)/G$ is exactly the cotangent bundle of $M/G$. Even when the action is not free, it makes sense to call the stack $[\mu^{-1}(0)/G]$ the cotangent stack of $[M/G]$. The "groupoid" resulting by performing the above construction on the action groupoid $G \times M$ indeed encodes this quotient.