Conductance of the mixture of two solutions with given resistance values
There is an excellent treatment of electrolytic conductivity in Wikipedia. However, the OP’s main question involved a specific equation, pertaining to the conductance of a mixture of equal volumes of two electrolyte solutions, and this was not specifically discussed in the Wikipedia article:

$$\text{conductance} = \frac{1}{2}\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right)$$

For simplicity, assume the electrolyte concentrations in the two solutions are in the “Goldilocks” zone: not so high that activities and fancy theory are needed, but not so low that the conductivity contribution due to water’s autoionization has to be taken into account. Neither solution is concentrated, so the ions just bounce around like they would in the unmixed solutions. This would certainly be incorrect in relatively concentrated solutions, but that is not the case at hand.

In this idealized scenario, a solution’s conductivity, i.e., specific conductance, is directly proportional to the mobile charge carrier (i.e., ion) concentrations in the solution. The conductance of a solution, denoted by $G,$ is directly proportional to the solution’s conductivity. The resistance of the solution, denoted by $R,$ is simply the reciprocal of the conductance and, likewise, the resistivity, i.e., specific resistance, is the reciprocal of the conductivity.

Now consider two electrolyte solutions that can be mixed without causing any chemical reaction. In particular, no precipitate forms, no gas is evolved, no complexes form, and so on. Measurements on the first solution provide values of $G_1$ and $R_1$, while measurements on the second solution provide values of $G_2$ and $R_2$. If equal volumes of the two solutions were well mixed, and a conductivity measurement was performed on the mixed solution, the conductivity of the mixed solution, $G_\mathrm{mix}$, would be

$$G_\mathrm{mix} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}(G_1 + G_2)$$

since each solution, when mixed with an equal volume of the other solution, had its constituent ion initial concentrations halved. But the total conductance is just the sum of the two conductances, $G_1 = 1/R_1$, and $G_2 = 1/R_2$, so

$$G_\mathrm{mix} = G_\mathrm{tot} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

Like all approximations, this one is only as good as the foundational assumptions upon which it is predicated.