When converting between a hydrated electrolyte amount in mass and in milliequivalents (meq), why are the water molecules taken into account?
When you want to obtain a solution of $\ce{CaCl2}$ in water, you can do it by dissolving anhydrous compound $\ce{CaCl2}$, or  the hydrated compound $\ce{CaCl2·2H2O}$. It does not matter which one to choose, provided you add enough water to compensate the difference and obtain at the end the wanted total volume.
If you want to obtain $1$ mL of a final solution that has to be $4$ meQ/ml, you must dissolve $4$ mEq of $\ce{CaCl2}$ or $4$ mEq of $\ce{CaCl2·2H2O}$ in $1$ mL . It does not matter which one you use. The only requirement is that the number of Equivalent has to be $4$ mEq. But the Equivalent weight  $\ce{CaCl2}$  is $111/2 = 55.5 $ g, and the equivalent weight of $\ce{CaCl2·6H2O}$ is $73.5$ g
To get the final solution with $\ce{CaCl2}$ you must choose to dissolve $4$ mEq $\ce{CaCl2}$, which weighs $\ce{(4·55.5}$ mg) = $222$ mg $\ce{CaCl2}$.
To get the same final solution with $\ce{CaCl2·6H2O}$ you must choose to dissolve $4$ mEq $\ce{CaCl2·6H2O}$, which weighs $\ce{(4·73.5}$ mg) = $294$ mg $\ce{CaCl2}$,
Do you understand ?