rotational dynamics - Where does the '2' Factor come from in the Coriolis Force?

Say we have a disk rotating at $\omega$, and an observer standing at radius $R$. The observer throws something of mass $m$ with radial velocity $v_r$, and the goal is to make this object go in a straight path along the radius. The Coriolis force says that we must exert a horizontal force of $2m\omega v_r$ - however, when I derive this myself, I get only $m\omega v_r$, as follows.

From an outsider's perspective, if we want the object to stay in a straight line, as the object proceeds towards the center, its horizontal velocity must change according to its position on the disk, such that $dv=\omega dr$. Thus the force on the object must be $m\frac{dv}{dt}=m\omega\frac{dr}{dt}=m\omega v_r$. Where did I go wrong in my derivation?