= Solution
Use primes for conformal-time <derivatives> in this question, reserving $V_{,\phi\phi}$ for a potential <derivative>. The <inflaton> background obeys $(a^2\bar\phi')'+a^4V_{,\phi}=0$, so the linear variation of its action vanishes up to boundary terms. With $\delta\phi=f/a$, the <conformal-time quadratic action for an inflaton perturbation> is initially
$$
S^{(2)}=\frac12\int d\tau\,d^3x\left[(f'-\mathcal Hf)^2-(\nabla f)^2-a^2V_{,\phi\phi}f^2\right],\qquad\mathcal H=a'/a.
$$
The cross term $-\mathcal H(f^2)'$ integrates to $\mathcal H'f^2$. Since $\mathcal H'+\mathcal H^2=a''/a$, this becomes
$$
\boxed{S^{(2)}=\frac12\int d\tau\,d^3x\left[f'^2-(\nabla f)^2+\left(\frac{a''}a-a^2V_{,\phi\phi}\right)f^2\right]}.
$$
Varying $f$ gives $f''-\nabla^2f+(a^2V_{,\phi\phi}-a''/a)f=0$. 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 yields
$$
\boxed{f_{\mathbf k}''+\left(k^2-\frac{a''}a\right)f_{\mathbf k}=0}.
$$
For <de Sitter spacetime>, $a=-1/(H\tau)$ with $\tau<0$, so $a''/a=2/\tau^2$. The rescaled field has a canonical kinetic term; $f$ rather than $\delta\phi$ is therefore the convenient oscillator variable.
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