Limiting molar conductivity of multiple solutes
I've found a answer for my question from this question.

The way to compute the conductivity of electrolyte with multiple ion types is given from the work Pawlowicz, Rich, ( 2008), Calculating the conductivity of natural waters, Limnol. Oceanogr. Methods, 6, doi:10.4319/lom.2008.6.489.

For general case consider the system, which consist of $N_+$ number of cation types, $N_-$ number of anion types and $N_{types} = N_+ + N_-$ total number of types, fully dissolved in solvent. This system can be viewed as a weighted sum of all possible pairwise combinations between cation and anion types. Then the conductivity $\sigma,\pu{[S m-1]}$ of the solution will be determined by:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} \Lambda_{eqm, \, ij}}$$
where $c_i^{\pm}$ - molar concentration of corresponding ion type, $\pu{[mol m-3]}=\pu{[mM]}$;

$z_i$ - valency of corresponding atom;

$c_i^{\pm} \cdot z_i$ - equivalent ionic concentration of corresponding ion type $\pu{[mol m-3]}=\pu{[mM]}$;

$\Lambda_{eqm, \, ij}$ - equivalent molar conductivity of binary subsystem of cation type $i$ and anion type $j$, $\pu{[S m^2 mol^{-1}]}$.

Equivalent ionic concentration $C_{eq}$ is defined as:
$$C_{eq} = \sum_i^{N_+}c_i^{+} \cdot z_i = \sum_j^{N_-}c_j^{-} \cdot z_j = \frac{1}{2}\sum_k^{N_{types}}c_k^{\pm} \cdot z_k$$

In general a binary solute dissociates according to formula:
$$\ce{A_{\nu^+}B_{\nu^-} -&gt; {\nu^+}A^{z^+ +} + {\nu^-}B^{{z^-}-}}$$
where $\nu^+$ and $\nu^-$ are coprime numbers and represent the moles of ions  for 1 mole of solute. It follows:
$$\nu^+ = z^-$$
$$\nu^- = z^+$$

Equivalent molar conductivity $\Lambda_{eqm}$ of single binary electrolyte is defined as:
$$\Lambda_{eqm} = \frac{\sigma}{c\nu_+z^+} = \frac{\sigma}{c\nu_-z^-}$$

For a binary solute, dissolved in water at normal pressure, $\pu{25 °C}$ and infinitely diluted, the law of the independent migration of ions can be written to find a limiting equivalent molar conductivity $\Lambda_{eqm}^{0}$:
$$\Lambda_{eqm}^{0} = (\frac{\nu_+}{z^+ } \lambda_{+}^{0} + \frac{\nu_-}{z^- }\lambda_{-}^{0}) = (\frac{z^-}{z^+ } \lambda_{+}^{0} + \frac{z^+}{z^- }\lambda_{-}^{0})$$
where $\lambda^{0}$ - limiting ionic conductivity of the ion, $\pu{[S m^2 mol^{-1}]}$.

Here interaction between ions is ignored.

Change of equivalent molar conductivity from changing of the solute equivalent concentration is determined by Debye-Hückel-Onsager equation of the form:
$$\Lambda_{eqm} = \Lambda_{eqm}^0 (1-A \sqrt{I}) - B \sqrt{I} =  \Lambda_{eqm}^0 K_A - K_B$$
where $A, \, B$ - Debye–Hückel–Onsager coefficients;

$I$ - stoichiometric ionic strength;
$$I = \frac{1}{2} \sum_{k=1}^{N_{types}}c_i z_i^2$$
$$ A = \frac{z^2eF^2}{3 \pi \eta}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$ 
$$ B = \frac{qz^3eF}{24 \pi \varepsilon RT}\left(\frac{2}{\varepsilon RT}\right)^{1/2}$$
where $\eta$ - viscosity of solvent, $\pu{[Pa s]}$;

$\varepsilon$ - dielectric permittivity of solvent;

$q$ - coefficient depends on $z^+/z^-$ of binary solute.

The expression for the conductivity of electrolyte (solvent - water at $\pu{25 °C}$ and normal pressure, ion pairng reduction factor is ignored ($\alpha_{ij}=1$ for all $i,j$)) will be:
$$\sigma = \sum_{i=1}^{N_+}\sum_{j=1}^{N_-}{\frac{c_i^+ z_i c_j^- z_j}{C_{eq}} (\frac{z_j^-}{z_i^+} \lambda_i^0 \, K_{A, \, ij} + \frac{z_i^+}{z_j^- } \lambda_j^0 \, K_{A, \, ij} - K_{B, \,ij})}$$

In my case I have 2 cations and 2 anions, their pairwise combination give 4 different solutes, which are assumed to be completely dissociate in water. Relation between solute molar concentration and its ions molar concentrations can be established.


$\ce{NaCl -&gt; Na+ + Cl-}$ 
$\ce{NaHCO3 -&gt; Na+ + HCO3-}$
$\ce{MgCl2 -&gt; Mg^2+ + 2Cl-}$ 
$\ce{Mg(HCO3)2 -&gt; Mg^2+ + 2HCO3-}$


Conductivity of my electrolyte will be:
\begin{align*}
\sigma &amp;= \frac{1}{[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]} \cdot \\
           &amp;\cdot \Bigl(\ce{[Na+]}  \ce{[Cl-]}       K_{A(\ce{NaCl})}        ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{NaCl})}}{K_{A(\ce{NaCl})}}      )+\\
           &amp;+             \ce{[Na+]}  \ce{[HCO3-]}  K_{A(\ce{NaHCO3})}  ( \lambda_{\ce{Na+}}^0 + \lambda_{\ce{HCO3-}}^0 - \frac{K_{B(\ce{NaHCO3})}}{K_{A(\ce{NaHCO3})}})+\\
           &amp;+             2\ce{[Mg^2+]}  \ce{[Cl-]}       K_{A(\ce{MgCl2})}        ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{Cl-}}^0      - \frac{K_{B(\ce{MgCl2})}}{K_{A(\ce{MgCl2})} }      )+\\
           &amp;+             2\ce{[Mg^2+]}  \ce{[HCO3-]}  K_{A(\ce{Mg(HCO3)2})}  ( \frac{1}{2}\lambda_{\ce{Mg^2+}}^0 + 2\lambda_{\ce{HCO3-}}^0 - \frac{ K_{B(\ce{Mg(HCO3)2})}} {K_{A(\ce{Mg(HCO3)2})}})
           \Bigr) 
\end{align*}

All the values here depend from given concentration of ions or from values found in tables. Hope it is correct, please, correct me if I'm wrong.