In the last step of Kjeldahl Method, is the reaction of ammonium to NaOH not taken into account?
The Kjeldahl method has two versions, using $\ce{HCl}$ as trapping solution or a weak acid like boric acid. Depending which one is used, the nitrogen content is either determined by a back titration with $\ce{NaOH}$ (your case) or a direct titration with $\ce{HCl}$.

The goal of the titration is to figure out how much ammonia (gas) has been trapped in the acid. The equation for the trapping step is:

$$\ce{NH3(g) + HCl(aq) -&gt; NH4+(aq) + Cl-(aq)}$$

So the trapping solution will get a bit less acidic because some of the $\ce{HCl}$ is neutralized by the weak base ammonia. To quantify the ammonia, you now titrate the remaining $\ce{HCl}$ with the strong base $\ce{NaOH}$:

$$\ce{HCl(aq) + NaOH(aq) -&gt; H2O(l) + Na(aq) + Cl-(aq)}$$

The solution is titrated to about $\mathrm{pH} = 5$ as endpoint (at the equivalence point, the strong acids and bases are neutralize, and $\ce{NH4Cl}$ solution remains). At $\mathrm{pH} = 5$, very little ammonium is deprotonated to form ammonia ($\mathrm{p}K_\mathrm{a}$ is about 9.3). In other words, ammonium/ammonia does not act as a buffer, so it does not affect the titration. You might even call it a spectator ion at this $\mathrm{pH}$.

After reaching the endpoint, the amount of $\ce{HCl}$ originally present in the trapping solution is equal to the sum of the amounts of $\ce{NaOH}$ and ammonia added. The overall reaction is (non-standard notation):

$$\ce{HCl(aq) + x NH3(g) + (1-x) NaOH(aq) -&gt;}$$
$$\ce{Cl-(aq) + x NH4+(aq) + (1-x) H2O(l) + (1-x) Na(aq)}$$

From that, you can calculate the amount of nitrogen in the sample.


  However, other than $\ce{HCl}$ there also exist an acid in the solution that can be titrated with $\ce{NaOH}$ which is the protonated form of $\ce{NH3}$.


As explained above, if the $\mathrm{pH}$ does not approach the $\mathrm{p}K_\mathrm{a}$ of ammonium, ammonium will not react with $\ce{NaOH}$ to an extent that matters. If you choose to end the titration when the solution reaches a $\mathrm{pH}$ of 11 (to give an example), however, the ammonium will be deprotonated to ammonia, invalidating the result. Worse, it might come out of solution, so it becomes a problem that can't be fixed.

Sources: 


http://www.brooklyn.cuny.edu/bc/ahp/SDKC/Chem/SD_KjeldahlMethod.html
J Martín, Fernández Sarria L and Agustín G. Asuero (2017) DOI: 10.5772/intechopen.68826


Wikipedia on the other hand I think got it wrong (https://en.wikipedia.org/wiki/Kjeldahl_method, retrieved before I edited it):


  Ammonium ion concentration in the standard acid solution, and thus the amount of nitrogen in the sample, is measured via titration. If boric acid (or some other weak acid) was used, direct acid-base titration is done with a strong acid of known concentration. $\ce{HCl}$ or $\ce{H2SO4}$ can be used. Indirect back titration is used instead if strong acids were used to make the standard acid solution: strong base of known concentration (like $\ce{NaOH}$) is first added in excess, and then the excess is titrated with a strong acid of known concentration. Titration of ammonia absorbed in boric acid solution thus has an advantage that only one standard solution is needed. One of the suitable indicators for these titration reactions is Tashiro's indicator.


They say to add an excess of $\ce{NaOH}$ and then titrate with $\ce{HCl}$. This would drive ammonia out of solution. Also, this is not a back titration but a double-back titration (which is not a thing, I think).