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:
$$\ce{A_{\nu^+}B_{\nu^-} -> {\nu^+}A^{z^+ +} + {\nu^-}B^{{z^-}-}}$$
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.
$\ce{NaCl -> Na+ + Cl-}$
$\ce{NaHCO3 -> Na+ + HCO3-}$
$\ce{MgCl2 -> Mg^2+ + 2Cl-}$
$\ce{Mg(HCO3)2 -> Mg^2+ + 2HCO3-}$
Conductivity of my electrolyte will be:
\begin{align*}
\sigma &= \frac{1}{[\ce{Na+}]+2 \cdot [\ce{Mg^2+}]} \cdot \\
&\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})}} )+\\
&+ \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})}})+\\
&+ 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})} } )+\\
&+ 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.
