Mechanism with hypohalite in haloform reaction
The mechanism of haloform reaction will also be almost similar in the case if we use hypohalites (here, it is $\ce{NaOCl}$).

In the first step, instead of $\ce{OH-}$, the acidic $\alpha$ proton will be taken up by $\ce{OCl-}$ (Remember, $\ce{HOCl}$ is a weak acid (but not as weak as water), so it's conjugate base i.e $\ce{OCl-}$ is a moderately strong base to take up a proton), and $\ce{HOCl }$ will be formed, which will be attacked by the enolate formed in the previous step, and $\ce{OH-}$ will leave to form the $\alpha$-haloketone. The reaction will proceed further two steps to form the tri-halo substituted ketone, and correspondingly $\ce{OH-}$ will be generated. These generated $\ce{OH-}$ can then attack the electrophilic carbonyl carbon give the haloform and the salt of the organic acid .

The mechanism is shown below,