How much lactose is there in milk (mechanism)?
Using the basics of organic chemistry, solutions of iodine and sodium hydroxide retain enough free $\ce{I2}$ to do other things. I would disocunt $\ce{NaIO}$ as a significant component of this process.

What if instead, we had exactly what your picture shows (except the last few steps are prevented - we'll see how in a few lines). You are worried abut the cyclic form, but the cyclic form and the acyclic form are in equilibrium in aqeous solution, especially if acid or base is present. Note that in the reactions equations below, I am using the notation $\ce{R}$ to refer to parts of the molecule that are not involved.


  Isomerization to the ene-diol


$$\ce{R-CH(OH)-CHO &lt;=&gt;[\ce{OH-}] R-C(OH)=CHOH}$$


  Oxidation by copper(II)


$$\ce{R-C(OH)=CHOH + 2Cu^2+ +3OH- -&gt; R-CH(OH)-CO2^- + 2Cu+ + 2H2O}$$

Now, if that copper(I) cations stumbles into a hydroxide anion, then it's all over since $\ce{Cu2O}$ is not soluble. However, it's iodine to the rescue! Iodine oxidizes $\ce{Cu+}$ back to $\ce{Cu^2+}$.

$$\ce{2Cu+ + I2 -&gt; 2Cu^2+ + 2I-}$$

These last two equations add to give the one you quote (only I left out the sodium spectator cations and simplified the carbohydrate).