I know if $f\in L^2(\mathbb R)$ is two times continuously differentiable, then we must have that the Fourier transform is integrable. Is there any more relaxed condition than this? For example if $f$ is continuously differentiable and $f^\prime $ is of bounded variation, will it imply that $f^\prime$ is integrable. Assume that all the functions are compactly supported.

2$\begingroup$ No, for example the FT of step function has decay $1/x$. (I assume that a hat is missing and you're asking if a BV function has integrable FT.) $\endgroup$ – Christian Remling Jun 27 '20 at 20:20
If $f'$ is of bounded variation, then $\hat{f}$ will be integrable. To see this, note that by your assumptions, $f''$ (in the sense of tempered distributions) is a finite complexvalued measure, so that $\widehat{f''}$ is bounded, meaning that $\hat{f}(\xi)\lesssim 1/(1+\xi^2)$.