You don't try to show that the map is transverse to $Z$, but rather take the iterated pullback $G_1\times_{G_0}G_1 \times_{G_0} \cdots \times_{G_0} G_1$.
For the case of $G_1\times_{G_0}G_1$ what you have is the pullback of $s$ along $t$, both submersions, so the projections are submersions, and hence the new map you are going to pull back is a submersion (being the composite of one of these projections and a source or target map). You iterate this and at each step the projection is a submersion, so the next step is do-able.
Edit:
Ah, I found the result I was looking for in Mackenzie's Lie groupoids and Lie algebroids in differential geometry (LMS lecture note series no. 124), namely Proposition III.1.17, on page 92. It says that for any Lie groupoid $\Omega \rightrightarrows B$ (there called a differentiable groupoid - for him Lie groupoids are a specialised notion) the inertia groupoid (there denoted $G\Omega$) is a sub-Lie groupoid, and the arrows of $G\Omega$ form a closed embedded submanifold of $\Omega$. In particular this result implies the source = target map $G\Omega \to B$ is a submersion.
However, in the newer General theory of Lie groupoids and Lie algebroids (LMS lecture note series no. 213), he corrects this, in a comment just after example 1.2.12, and only claims it for locally trivial Lie groupoids, which was probably what he was thinking of in the earlier book.
