Let $f: \mathbb R \rightarrow \mathbb R$ be a continuous function with compact support. Assume further that $f$ is smooth, expect for at the origin where it has a discontinuous derivative $$ \Delta := \lim_{\epsilon \rightarrow 0^+} ( f'(\epsilon)-f'(-\epsilon)) \neq 0. $$ Define $$ g(x) := \int_{-\infty}^{\infty} dk \int_{-\infty}^{\infty} dy~ k^2 e^{ik(x-y)} f(y) $$ and $$ I := \int_{-\infty}^{\infty} dx \, g(x). $$ Assuming that I can freely interchange the integrals, in $I$ I can pull in the $x$ integral, which gives a delta distribution $\delta(k)$ and therefore $$ I = 0. $$ On the other hand, let us calculate $I$ by partially integrating twice $$ I = - \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dk \int_{-\infty}^{\infty} dy~ e^{ik(x-y)} f''(y) = - 2\pi \Big ( \int_{-\infty}^0 dx\, f''(x) + \int_0^{\infty} dx\, f''(x) \Big ) = - 2\pi \Delta. $$ (I am aware that this is probably illegal, since $f'$ is not absolutely continuous, right?) This suggests that $g$ has an additional delta function term $$ - 2\pi \Delta \, \delta(x), $$ which does not appear in the first calculation.
This is a contradiction, so obviously I have made an error in one (or both) of the calculations.
Is $g$ well-defined for the function $f$ at hand? Is it sensible to write $$ g(x) = - 2\pi f''(x) $$ in the sense of distributions? So that for a suitable test function $\phi$, we have $$ \int_{-\infty}^{\infty} dx~ g(x) \phi(x) = 2\pi \Delta \, \phi(0) + 2\pi \int_{-\infty}^{\infty} dx~ f'(x) \phi'(x). $$ Should the $\Delta$ term be there or not?
The identity $ \int_{-\infty}^{\infty} k^2 \hat{f}(k) e^{ikx} \, dk = 2\pi \, f''(-x) $ is only valid if $f'$ is continuous. Otherwise one must replace $f''$ with the distributional second derivative.
Let $f$ be as in the question and define $\partial^2 f$ by $$ \partial^2 f(x) = \begin{cases} f''(x) & (x \neq 0) \\ 0 & (x = 0) \end{cases} $$ and let us denote the distributional second derivative with $D^2 f.$ Then we have $$ D^2 f = \partial^2 f + (f'(0^+)-f'(0^-)) \, \delta. $$ Now, when we integrate this we get $$ \int_{-\infty}^{\infty} D^2 f(x) \, dx = \int_{-\infty}^{0} \partial^2 f(x) \, dx + \int_{0}^{\infty} \partial^2 f(x) \, dx + \int_{-\infty}^{\infty} (f'(0^+)-f'(0^-)) \, \delta(x) \, dx = (f'(0^-)-f'(-\infty)) + (f'(\infty)- f'(0^+)) + (f'(0^+)-f'(0^-)) = 0. $$