 Solution

Use primes for conformal-time derivatives in this question, reserving $V_{,\varphi \varphi }$ for a potential derivative. The inflaton background obeys $(a^2{\bar{\varphi }}^{\prime }{)}^{\prime }+a^4{V}_{,\varphi }=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta \varphi =f/a$, the conformal-time quadratic action for an inflaton perturbation is initiallyThe cross term $-\mathcal{H}(f^2{)}^{\prime }$ integrates to ${\mathcal{H}}^{\prime }{f}^2$. Since ${\mathcal{H}}^{\prime }+{\mathcal{H}}^2=a^{\prime \prime }/a$, this becomesVarying $f$ gives $f^{\prime \prime }-{\nabla }^{2}f+(a^2{V}_{,\varphi \varphi }-a^{\prime \prime }/a)f=0$. The symmetric Fourier transform convention used in the question turns $-{\nabla }^2$ into $k^2$. Dropping the small mass term under the stated slow-roll inflation assumption yieldsFor de Sitter spacetime, $a=-1/(H\tau )$ with $\tau <0$, so $a^{\prime \prime }/a=2/{\tau }^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta \varphi$ is therefore the convenient oscillator variable.