Are Legendre transforms of non-convex functions useful?
(Edit added years later: I want to make clear that J.R.'s answer - the accepted one - is completely correct if we're using the Legendre-Fenchel transform, which is the one people generally use. The Legendre-Fenchel transform is defined in terms of the supremum and can be applied to non-differentiable functions. I believe that Legendre originally defined his transform in terms of explicitly setting the derivative to zero instead, which is what I'm doing below. So this answer is about the original Legendre transform, rather than the usual Legendre-Fenchel one.)
I would like to propose a different answer to the second part of your question. Let us examine a simple example of a non-convex function, $A(x) = x^2-x^4$, which of course looks like this:
(plots are from Wolfram|Alpha.)
Let us define $y=\frac{dA}{dx}=2x-4x^3$ and consider the Legendre transform $B = xy-A(x) = x^2 - 3x^4$. Solving explicitly to get $B$ as a function of $y$ involves solving a cubic equation and doesn't seem to lead anywhere nice, but we can easily do a parametric plot of $B(x)$ against $y(x)$ to see what it would look like. The result is this: ($B$ is the vertical axis and $y$ is the horizontal axis)
I believe that Legendre transforms of functions with a single non-convex region will always be qualitatively similar to this. In general the Legendre transform of such a function will be a multiple-valued function with this distinctive "swallowtail" shape. It won't always be symmetrical of course, but there will always be two points where the curve reverses direction and one point where it crosses itself. The direction of curvature of the three line segments will also always follow the pattern of this plot.
The reason the curve reverses direction is simply that $y$ is the slope of the $A(x)$ curve. For a convex function, $\frac{dy}{dx} = \frac{d^2 A}{dx^2}$ is always negative, so $y$ changes monotonically with $x$; but for a non-convex function there is a region where $\frac{dy}{dx} = \frac{d^2 A}{dx^2}>0$, corresponding to a reversal in the slope of $y(x)$. Indeed we can see this non-monotonicity by plotting $y=2x-4x^3$ against $x$:
If there are multiple concave regions then there will be multiple reversals of the sign of $dy/dx$, corresponding to multiple swallowtails in the Legendre transform. I think these multiple swallowtails can be sort of nested inside one another.
If one replaces the non-convex function with its convex hull then the "swallowtail" will disappear, with the point where the curve crosses itself becoming a "kink" in the curve, where the second derivative becomes infinite. This results in a single-valued convex function.
As for whether this interpretation is useful, I'm not sure but I think so. It seems related to the swallowtail catastrophe (although I don't know much about catastrophe theory), and it also seems related to first-order phase transitions in statistical mechanics. These can be seen as being due to a non-convexity in the entropy (which then gets replaced by its convex hull). This is equivalent to a swallowtail in the free energy being replaced by a kink, since the two are Legendre transforms of one another. I've never seen this swallowtail curve depicted in a thermodynamics text, but it's not uncommon to see plots corresponding to the $y(x)$ plot above.
I can answer your second question.
Recall that by convex duality, applying the Legendre transform twice on a convex (and coercive) function gives you back the original function. Now even if you plug in a non-convex function $W$ instead, the Legendre transform $W^*$ will be convex and coercive (one can see that by the definition) and so is $V:=(W^*)^*$. So clearly the convex duality will not carry over to the non-convex case. However one can show (under appropriate assumptions of course) that $V$ is the convexification or convex hull of $W$, that is roughly (here $W:\mathbb{R}\rightarrow\mathbb{R}$) the function whose graph is the boundary curve of the convex hull of the set $\{(x,y)\in\mathbb{R}^2\,|\,y\ge f(x)\}$. By convex duality we conclude that $V^*=((W^*)^*)^*=W^*$, so the Legendre transform of a non-convex function is the Legendre transform of it's convexification.