Boiling point elevation with multiple solutes
When there is more than $1$ solute in the solution, the total elevation in boiling point ($\Delta T_\mathrm{b,total}$) or the depression in freezing point ($\Delta T_\mathrm{f,total}$) is defined as the sum of each individual value for each type of solute.
What is Van't Hoff factor? According to Chem Libretexts, the Van't Hoff factor is defined as:

$$i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \tag{1} \label{1}$$

Now, why does this work? Let's take an example to see how it works.
We take a simple case as what you have mentioned in the question:

Find the elevation in boiling point of a solution of $\pu{1 m}$ glucose and $\pu{2 m}$ $\ce{NaCl}$ [$100$% dissociated] in water.
$\mathrm k_\mathrm {b,water}= \pu{0.52 ^\circ C kg mol-1}$

We can do this in two ways. One, where we take $i = 2$ for $\ce{NaCl}$. Second, we state that $\ce{NaCl}$ dissociates into $\ce{Na+}$ and $\ce{Cl-}$ and take the three types of solutes to be present instead of two [going from basic principles].
Method $1$
Here we would have two different solute particles. Finding the value of $\Delta T_\mathrm{b}$ for both glucose and $\ce{NaCl}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &amp;= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,NaCl} \\
&amp;= \mathrm{k_b}[1\cdot1 \,+2\cdot2] \\
&amp;= 5\cdot\mathrm{k_b} \\
\end{align}
Method $2$
Here we would have three different solute particles. Here, $\ce{Na+}$and $\ce{Cl-}$ would have a value of $i=1$ and their individual molality would be identical to that of $\ce{NaCl}$ since the salt was $100$% dissociated in solution
Finding the value of $\Delta T_\mathrm{b}$ for glucose, $\ce{Na+}$ and $\ce{Cl-}$   using the formula $\Delta T_\mathrm{b} = i\,\mathrm{k_b}\,\mathrm m$ we get the net value to be:
\begin{align}
\Delta T_\mathrm{b,total} &amp;= \Delta T_\mathrm{b,glucose}+\Delta T_\mathrm{b,Cl^-} +\Delta T_\mathrm{b,Na^+} \\
&amp;= \mathrm{k_b}[1\cdot1 \,+1\cdot2 + 1\cdot2] \\
&amp;= 5\cdot\mathrm{k_b} \\
\end{align}
Therefore, both methods yield the same answer. The second method is actually the derivation for value of $i$ for a dissociating salt.
Now, you can apply $i_\text{total}=i_1+i_2$ only when the concentrations of the given solutes are the same. The proof is similar to how the second method proves the first above. In such a scenario, you can assume them to be one solute since colligative properties don't care about how many types of solute there are but rather care only about the amount of solute particles present in the solution.