If it works at all, it would only be when all three bodies are aligned. If the planet-moon line is perpendicular to the planet-sun line, there's no cancellation of the stars gravitational field between the planet and moon.
It works, but moves the low-G region closer to the star (and likely towards the moon, so its gravity can offset the star's). You calculate the low-G point between the planet and moon, then move it
towards the star slightly to counteract the star's gravity. This of it this way: if you drop something on the moon, does it fall into the Sun, or onto the Moon?
Pirate points for planet-star exist in Battletech (Strategic Operations, p133 - the Transit Time chart, right-hand column: Transit to Sol-Planet L1 point). Even though Terra's jump limit for Sol is ~10 AU, there is a pirate point between Terra and Sol. That distance is obviously less than 10 AU, and obviously within Sol's jump limit. That point is still a three-body problem, as Luna provides a small amount of gravity, even at that distance.
By a similar process there is a pirate point between Terra and Luna, so that the Sun's gravity barely changes at that distance, while the angled gravity sources from Terra and Luna can rise sufficiently to reduce the net gravity to be low enough to allow a KF jump to occur. It is a difficult target (as listed in the rules with a 4 pt penalty to the target number), but it still exists.
A rough example would be imagining a 3-body problem, where there are a pair of twin planets orbiting a common center, and the local star. If you can pick a point so there is a gross of .01G pulling at a direction of 10 o'clock (from planet A), a gross value of .01G pulling from 2 o'clock (planet B), and a gross of .01G pulling from 6 o'clock (the local star), the three sources of gravity effectively null each other out at that specific spot. It won't last that long as the two planet rotate about their common center, but for a brief period of time the gravity at that location will be low enough.
Edit - had to change one word
