We consider the function
$$f (x) = \begin{cases} 1+x & -1\leq x\leq 0\\ 1-x & 0\leq x\leq 1 \\ 0 & \text{otherwise} \end{cases}$$
and the formula
$$F(f'')(\xi)=-\xi^2F(f)(\xi)$$
I have used the fact that $f$ is twice-differentiable almost everywhere (it has a problem at $x=0,-1$ and $1$) and that this wouldn't affect the integral in the Fourier transform to get that the Fourier transform of $f$ is $0$ on all points that are not $0$, which is false.
What would be the error and how would be the correct way to use that formula?
The formula $F(f'')(\xi)=-\xi^2F(f)(\xi)$ is valid when the derivative $f''$ is the distributional derivative of $f$. When $f\in C^1$, the classical derivative $f'$ represents the distributional derivative. When $f$ is absolutely continuous, $f'$ exists a.e. and still represents the distributional derivative. Beyond that is the danger area; "almost everywhere" is not enough, as a single point may be the support of a distribution.
Here $f$ is Lipschitz continuous, hence absolutely continuous. So $$f' (x) = \begin{cases} 1 \quad & -1 < x< 0\\ -1 & 0< x < 1 \\ 0 & |x|>1 \end{cases}$$ correctly represents the distributional derivative.
However, $f'$ is not continuous. Its derivative is $0$ a.e., but since $f'$ has jumps, there's a singular part of the derivative. Specifically,
$$f '' (x) = \delta (x+1) - 2 \delta (x) + \delta (x-1)$$ as Rodrigo de Azevedo pointed out. To see why, notice that $$f ' (x) = H (x+1) - 2H(x) + H(x-1)$$ where $H$ is the Heaviside function, and recall that $H'=\delta$, the Dirac delta.