Here is one limit on the curvature needed. It is not optimal, but it is explicit, and probably a good starting point for numerical or theoretical optimization.

**Proposition:** Consider the vertices with their convex hull and its sides; let $h$ be the minimum over all sides of 
$$\frac{s_j}{\tan\left(\frac\pi4+\frac{t_j}2\right)
+\tan\left(\frac\pi4+\frac{t’_j}2\right)}$$
where $s_j$ is the length of the side and $t_j$, $t’_j$ are half the angles at its vertices.
Then there is a path through the vertices whose maximal curvature is 
$$\kappa_{\max} = \frac{1-\sin(\min t_i)}{h}$$
where again each $t_i$ is half of a vertex angle.

The path will go through each vertex and a point or line segment at a height $h$ above each side.

**Proof:** For consecutive vertices $UVW$, take a point $O$ at distance $h/(1-\sin t)$ from $V$ along the bisector of angle $UVW$. We will construct a path which includes the circle arc which is centered at $O$, goes through $V$, and is bounded by the two points $P$ and $Q$ with angular distance of $\frac\pi2-t$ from $V$.

For these points, $OP \perp UV$, $O$ is at distance $(h \sin t)/(1-\sin t)$ from $UV$, $P$ is at distance $h$ from $UV$ and the arc covers a distance of $(h \cos t)/(1-\sin t)$ or $h\tan(\frac\pi4+\frac{t}2)$ along $UV$. Similar claims hold for $OQ$ and $VW$.

Now we construct the path as the combination of all the arcs $PVQ$ and any needed lines connecting them, where all the lines would be at distance $h$ from the convex hull. The definition of $h$ ensures that none of the circle arcs overlap. Its curvature in each arc is $(1-\sin t)/h$, and its curvature in each line is $0$, so the maximal curvature is as stated. $\square$

**Example:** Let the vertices be $(\pm2,\pm1)$ and $(\pm3,0)$. Then each side length is $\sqrt{2}$ or $4$, and each vertex half-angle is $\pi/4$ or $3\pi/8$. So $h$ is the minimum of
$$\frac{\sqrt{2}}{\tan\frac{3\pi}8+\tan\frac{7\pi}{16}}\sim0.19$$
$$\frac{4}{2\tan\frac{7\pi}{16}}\sim0.40$$
and $h\sim 0.19$, with a maximal curvature of $(1-\sin\frac\pi4)/h\sim 1.54$. 

The apparent minimum possible maximum curvature for this example is $1$, achieved by the union of two vertical semicircles and two horizontal lines. This path goes to a height of $1-1/\sqrt{2}\sim 0.29$ above the four diagonal sides, and to a height of $0$ above the two horizontal sides. So here the construction in the proposition gets roughly the right average height, and an optimization might proceed by increasing and decreasing the height of the path in different places.