I need to report MoS2 values from Mo and Sulfur numbers
As the molecular mass of molybdenum is $95.95~\mathrm{g/mol} \approx 96~\mathrm{g/mol}$ and the molecular mass of sulfur is $32.06~\mathrm{g/mol} \approx 32~\mathrm{g/mol}$ one can construct a ratio between both molecular masses that in turn can be solved easily.
$$\frac{96~a}{32~b}=\frac{5.2}{5.3}=0.98\approx1$$
Which gives us after the multiplication with 32/96 on both sides
$$\frac{a}{b}=1\cdot\frac{32}{96}=\frac{1}{3}$$
So customer 1's determined "compound" is not $\ce{MoS2}$ but more $\ce{MoS3}$. Now I would guess that they don't analyze the copper ore for the compound $\ce{MoS2}$ but for both elements $\ce{Mo}$ and $\ce{S}$ separately which is why they came up with $5.2~\%~\ce{Mo}$ and $5.3~\%~\ce{S}$.

In my opinion, you should ask customer 2 to split up it's $12.34~\%~\ce{MoS2}$ into the percentages of each element and compare them with $5.2~\%$ and $5.3~\%$.

Using the ratio for $\ce{MoS2}$ which is $60:40$ one would estimate those $12.34~\%$ to split up into $7.40~\%~\ce{Mo}$ and $4.94~\%~\ce{S}$ or into $6.16~\%~\ce{Mo}$ and $6.18\%~\ce{S}$ if it is more like $\ce{MoS3}$.

Or to answer your question ... probably you can simply add both values to give you $10.5~\%$ which nonetheless will not be equal to $\ce{MoS2}$ based upon the data you are given!