I will give an example as to why composition of bibundles cannot simply be done as pullback, as well as the relation to perhaps more familiar geometric constructions.

Firstly, note that if $M$ is a manifold and $\mathbf{B}G$ is a one-object Lie groupoid (where the automorphisms of the one object are the Lie group $G$), a bibundle from $M$ to $\mathbf{B}G$ is precisely the same thing as a principal $G$-bundle on $M$. Now consider a homomorphism $\phi\colon G\to H$, and the corresponding functor $\mathbf{B}G\to \mathbf{B}H$. This gives a bibundle from $\mathbf{B}G$ to $\mathbf{B}H$, namely $u(H) \to \ast = Obj(\mathbf{B}G)$, where $u(H)$ is the underlying manifold of the Lie group $H$. The left $\mathbf{B}G$-action is trivial, and the right $\mathbf{B}H$-action (which is the same as an $H$-action) is by multiplication. The naive "composition" of bibundles should give a bibundle from $M$ to $\mathbf{B}H$, which is just a principal $H$-bundle on $M$, but what it actually gives is just $P\times H \to M$, which is the composite of the projection on $P$ and the given map $P\to M$.

The correct composite should be the principal bundle you get by changing the structure group along the given homomorphism $G\to H$, which is the quotient $(P\times H)/G$ by the action of $G$ on $P\times H$ as $(p,h) \mapsto (pg,\phi(g)^{-1}h)$. Equivalently, one can set up an equivalence relation on $P\times H$ so that $(pg,h) \simeq (p,\phi(g)h)$. This generalises directly to the case when you replace $\mathbf{B}G\to \mathbf{B}H$ by some arbitrary functor $X\to Y$ of Lie groupoids (and compose with the bibundle it gives rise to), and with only a little more work when you replace the manifold $M$ by a general Lie groupoid. At that point, you probably will be comfortable with the general case.

I will give an example as to why composition of bibundles cannot simply be done as pullback, as well as the relation to perhaps more familiar geometric constructions.

Firstly, note that if $M$ is a manifold and $\mathbf{B}G$ is a one-object Lie groupoid (where the automorphisms of the one object are the Lie group $G$), a bibundle from $M$ to $\mathbf{B}G$ is precisely the same thing as a principal $G$-bundle on $M$. Now consider a homomorphism $\phi\colon G\to H$, and the corresponding functor $\mathbf{B}G\to \mathbf{B}H$. This gives a bibundle from $\mathbf{B}G$ to $\mathbf{B}H$, namely $u(H) \to \ast = Obj(\mathbf{B}G)$, where $u(H)$ is the underlying manifold of the Lie group $H$. The left $\mathbf{B}G$-action is trivial Edit the left action of $G$ on $u(H)$ via $\phi$ [end edit], and the right $\mathbf{B}H$-action (which is the same as an $H$-action) is by multiplication. The naive "composition" of bibundles should give a bibundle from $M$ to $\mathbf{B}H$, which is just a principal $H$-bundle on $M$, but what it actually gives is just $P\times H \to M$, which is the composite of the projection on $P$ and the given map $P\to M$.

The correct composite should be the principal bundle you get by changing the structure group along the given homomorphism $G\to H$, which is the quotient $(P\times H)/G$ by the action of $G$ on $P\times H$ as $(p,h) \mapsto (pg,\phi(g)^{-1}h)$. Equivalently, one can set up an equivalence relation on $P\times H$ so that $(pg,h) \simeq (p,\phi(g)h)$. This generalises directly to the case when you replace $\mathbf{B}G\to \mathbf{B}H$ by some arbitrary functor $X\to Y$ of Lie groupoids (and compose with the bibundle it gives rise to), and with only a little more work when you replace the manifold $M$ by a general Lie groupoid. At that point, you probably will be comfortable with the general case.