Solution:

Suppose vertex $A$ is that marked $*$ at the bottom left. Without loss of generality, $B$ is in a $60$ degree sector as shown. Assume the edges have unit length. The vertices can be partitioned into two sets (marked $\circ$ and $\cdot$ in the diagram). Each set forms a skewed lattice with axes at $60$ degrees. Any path must alternate between the two lattices.
If $B$ is on the same lattice as $A$, then we can give $B$ coordinates $(m, n)$ relative to $A$ and the shortest path from $A$ to $B$ must move $m$ units east and $n$ units east of north. The shortest path between a lattice point and the next lattice point east is evidently one edge in direction $3$ followed by one edge in direction $2$. Similarly, the shortest path between a lattice point and the next lattice point east of north is one edge in direction $1$, followed by one edge in direction $2$. So a shortest path from $A$ to $B$ must have $\{m + n\}$ edges in direction $2$.
$B$ is a distance $\sqrt{3}(m + n/2)$ east of $A$ and a distance $3n/2$ north of $A$, so $AB^2 = 3m^2 + 3mn + 3n^2 < 4m^2 + 8mn + 4n^2 = 4(m + n)^2$. So in this case the bug must travel more than $AB/2$ in direction $2$.
Now suppose $B$ is on the other lattice. Let $C$ be the lattice point immediately north of $A$ and $D$ the lattice point in direction $3$ from $A$. Then a shortest path from $A$ to $B$ must either be $A$ to $C$ and then a shortest path from $C$ to $B$, or $A$ to $D$ and then a shortest path from $D$ to $B$. Take $B$ to have coordinates $(m, n)$ relative to $C$ or $D$.
In the first case, $AB^2 = (\sqrt{3}(m + n/2))^2 + (3n/2 + 1)^2 = 3m^2 + 3mn + 3n^2 + 3n + 1$ and a shortest path has $m + n$ units in direction $2$. But $4(m + n)^2 > 3m^2 + 3mn + 3n^2 + 3n + 1$, if $m^2 + n^2 + 5mn > 3n + 1$, which is true for $m, n$ at least $1$. If $m = 0$ and $n = 1$, then a shortest path has $2$ units in direction $1$ and $AB = \sqrt{7} < 4$. If $m = 1$ and $n = 0$, then $AB = 2$ and a shortest path has $1$ unit in each direction. So in this case (the only one so far) we have equality.
It remains to consider the case where the path starts out towards $D$. In this case $AB^2 = (\sqrt{3}(m + n/2) + \sqrt{3}/2)^2 + (3n/2 - 1/2)^2 = 3m^2 + 3mn + 3n^2 + 3m + 1$ and a path has $m + n$ units in direction $2$. But $4(m + n)^2 > 3m^2 + 3mn + 3n^2 + 3m + 1$ for $m^2 + n^2 + 5mn > 3m + 1$, which is true for $m, n$ at least $1$. If $m = 1$, $n = 0$, then a shortest path has $2$ units in direction $3$ and $AB = \sqrt{7} < 4$. Finally, if $m = 0$ and $n = 1$, then a shortest path has $1$ unit in each direction and $AB = 2$.
Thus the answer to the final question is $3$, because the only cases where the bug travels exactly $AB/2$ in one direction are where it goes to the opposite vertex of a hexagon it is on.