It simply refers to electrons in orbitals with the same sign of $m_l$.
The quick and easy way to determine the ground-state term symbol of a lanthanide is to fill the 4f orbitals from most positive $m_l$ to most negative. This is in order to get the largest possible value of $M_L$:
$$M_L = \sum m_l$$
This is not valid by quantum mechanical arguments due to the indistinguishability of the electrons. However, it always leads to the correct ground-state term symbol because of Hund's second rule, which essentially states that the term with the largest value of $L$ is the most stable (after the first rule has been taken into account). Since the value of the projection quantum number $M_L$ ranges from $-L, -L+1, \ldots, +L$, a ${}^{2S+1}L$ term necessarily contains a term with $M_L = +L$. Therefore, by maximising the value of $M_L$, you will automatically generate the term with the largest possible $L$.
For example, consider the $\mathrm{f^3}$ ion $\ce{Nd^3+}$. You would want to "fill the orbitals" in this way:
where you have first and foremost put the electrons all with the same spin in order to satisfy Hund's first rule, and then filled them in decreasing order of $m_l$ in order to satisfy Hund's second rule.
The classical argument behind Hund's second rule is that electrons that have the same sign of angular momentum "collide" into each other less and as a consequence repel each other less. It is like two people running around a track; if they ran in the same direction, they would bump into each other less than if they ran in opposite directions.
Therefore, removal of an electron with "like rotation" (as you would if you were to ionise $\ce{Nd^3+}$ to $\ce{Nd^4+}$) is marginally more difficult than expected. Do not imagine it as a huge effect. It is very small and only leads to minor deviations from the ionisation energy trends (quarter-shell and three-quarter-shell effects).
