In what case would bi-directional BFS have a worse asymptotic run time than regular BFS?
When attempting to find a path from a source vertex $s$ to a target vertex $t$, with both BFS and bi-directional BFS, if you assume a fairly steady branching factor $b$, the number of possibilities checked (and hence the running time) is dependent on the depth. For BFS the depth is the distance from the source to the target, giving $b^{k}$ vertices visited, for bi-directional BFS the two searches meet in the middle, so it's two lots of half-distance searches - $2\cdot b^{k/2} = 2\cdot\sqrt{b^{k}}$, which is clearly better.

So to get a worse time, you need a route from $s$ to $t$ that traverses very few vertices, but the route from $t$ to $s$ would visit many vertices. Essentially you want an asymmetric branching factor.

The family of star graphs would be sufficient for this. A star graph has a single central vertex, and all other vertices are connected directly to it, and no other vertices. So if you have $s$ as one of the pendent vertices, and $t$ as the central vertex, normal BFS would take constant time, and bi-directional BFS would take linear time. You can also invent more complicated examples that share this property of a low branching factor going one way, but a high one coming the other (up to a point).