The examples pointed out by Ulrich Pennig are the bundles $U(n-1) \to U(n) \to S^{2n-1}$, $SU(n-1) \to SU(n) \to S^{2n-1}$, and $Sp(n-1) \to Sp(n) \to S^{4n-1}$ so one doesn't have to look far to find such examples. Imagine what Lie group theory would be like if all these bundles were trivial, so all these groups would be products of odd-dimensional spheres!

After all the pairs of boundary arcs have been identified one has a model surface $M(a,b,c,d)$ with $d=0$, so all that remains to show is that if $b$ or $c$ is nonzero then all handles and cross-handles can be replaced by cross-caps, which isn't hard for the standard models. The operation of forming a connected sum plays no role in this proof, nor does the induction step involve attaching handles, cross-handles, or cross-caps in the interior of a surface.

Once one specifies the two oriented arcs to be identified, the choice of an orientation-preserving homeomorphism identifying them doesn't matter since each orientation-preserving homeomorphism of one of the arcs extends to a homeomorphism of the model surface (before identification) using a collar neighborhood of the boundary. (continued)

With the Zip proof the problems you are worrying about do not actually arise. To begin, fix a set of standard model surfaces $M(a,b,c,d)$ with $a$ handles (tubes), $b$ cross-handles, $c$ cross-caps, and $d$ boundary circles. One just has to show that if one takes a collection of disjoint model surfaces and identifies two oriented boundary arcs, then the result is again homeomorphic to a collection of disjoint model surfaces. (continued)

There is a discussion somewhat along these lines in my algebraic topology book (available online) in Section 3.H, pages 327-336.

@Akerbeltz: I think you are understanding this correctly. The twelve missing tetrahedra are grouped into six pairs with the two tetrahedra of each pair interchanged by reflecting the $I$ factor of $\Delta_2\times I$ across its midpoint. This means that after projecting $\Delta_2\times I$ onto $\Delta_2$ the two tetrahedra in each pair cancel algebraically. The inductive construction implies that there is a similar cancelation in all dimensions.

@Akerbeltz: If you form the cone on a triangulation of $\Delta_p\times \partial I \cup \partial \Delta_p \times I$ then the result is a triangulation of $\Delta_p\times I$. In your picture with 16 tetrahedra the underlying topological space is not homeomorphic to $\Delta_2\times I$ so it cannot give a triangulation of $\Delta_2\times I$. (A horizontal plane halfway between the top and bottom of your figure intersects the figure in something one-dimensional instead of two-dimensional.)

@Akerbeltz: Nice graphic at the link you provided, but something seems to be missing. There are 16 tetrahedra shown but there should be 28.