Lugol's iodine 5 % aqueous solution
A Lugol's iodine solution is an aqueous solution of elemental iodine and potassium iodide. This solution may be prepared in different strengths, typically $2{-}15~\%$, while the strength is determined by the content of elemental iodine (see this table from the Wikipedia article, but note that the first column's title is a bit misleading — the leftmost column lists the different strengths of Lugol's solution, and the other columns list the amount of the each constituent (iodine, iodide, total) in a solution of that strength).

Now, this is in essence an iodine solution, but it is prepared from iodine and potassium iodide due to solubility issues of elemental iodine. See this answer on Chemistry.SE to the question Preparation of iodine solution for a simple and concise explanation.

For the calculation of the amount of iodine and potassium iodide in solution, let us take $5~\%$ Lugol's solution, which is prepared from 5 grams of iodine and 7.5 grams of potassium iodide in $\pu{100 mL}$.

If we have $\pu{5 g}\ (\pu{5000 mg})$ of iodine and $\pu{7.5 g}\ (\pu{7500 mg})$ of potassium iodide in $\pu{100 mL}$, then after dividing by $100$, we have $\pu{50 mg}$ of iodine and $\pu{75 mg}$ of potassium iodide in $\pu{1 mL}$.

Since the commonly accepted volume of one drop is $1/20\ (5~\%)$ of $\pu{1 mL}$, then by multiplying $\pu{50 mg}$ and $\pu{75 mg}$ by $0.05$, we get $\pu{2.5 mg}$ of iodine and $\pu{3.75 mg}$ of potassium iodide in one drop of the solution.

For better clarity, here is the calculation again for I2 content:

$$\frac{\pu{5 g}}{\pu{100 ml}}\cdot \frac{\pu{1000 mg}}{\pu{1 g}} = \frac{\pu{50 mg}}{\pu{1 mL}}  \\ \Longrightarrow  \frac{\pu{50 mg}}{\pu{1 mL}}\cdot \frac{\pu{1  mL}}{\pu{20 drops}} = \frac{\pu{2.5  mg}}{\pu{1 drop}}$$

Do the same for $\pu{7.5 g}$ of potassium iodide to get $\pu{3.75 mg/drop}$.

A word about the actual iodine content of the solution following Jan's comment:

In principle, masses can always be added up (unlike volumes, which should not be added up, due to possible interactions between certain solvents — see here).
Therefore, we can say that the total mass of solutes in one drop is $$\pu{2.5mg}+\pu{3.75 mg} = \pu{6.25 mg}$$ ($\pu{12.5 mg}$ in two drops, per OP question). However, we cannot say that two drops contain $\pu{12.5 mg}$ of iodine, because the solution contains both elemental iodine (I2) and potassium iodide (KI), which are two different constituents. Since KI's role in the solution is to facilitate dissolution of elemental iodine molecules (see description of chemical properties, including equations, here), iodide content is irrelevant and the solution's strength reflects only iodine's content.

In other words, $\pu{12.5 mg}$ is the total quantity of solutes in two drops of Lugol's solution, but this quantity has no practical meaning, since the active ingredient is iodine, and its amount in 2 drops of $5~\%$ Lugol's solution is $\pu{5 mg}$.