This is incorrect, let $f = 1$, then an ideal maximal with this property is a maximal ideal. But no maximal ideal has a direct successor.

One would have to assume something about $f$ so that $f+P$ is not a unit. But I'm not sure what the condition would be for now

The following does hold:

\begin{align*} P \text{ is a lower ideal} & \iff \\ P \text{ is not maximal, and there is }&f\text{ such that }P\text{ is maximal with respect to ideals }I\text{ with }I\cap \{f^n\} = \emptyset\end{align*}

One direction of 14D.1 is also only true when $P$ is not maximal