Could you explain that a little? I'm afraid I don't know what you're saying at all. :|
OK, now you're going to regret asking such things of a former meteorologist and freshman-level physics TA.
An orbit around a celestial object is basically falling forever and always missing that thing.
Let's start with the basics: An object in motion stays in motion, unless acted upon by an external force. And by motion, we mean a straight line, neither speeding up nor slowing down. Such things only happen in the abstract world of simplified physics, which is just a subset of mathematics. Real life is messy, as the Chinese Tiangong-1 space station is demonstrating today.
So, first problem: how to put something in an orbit that doesn't involve crashing. Top problem is gravity; it's always there. The acceleration that any object imposes on another object is equal to the universal gravitational constant (denoted usually as capital G) times the mass of that object divided by the square of the distance between those objects. In simplified physics, you calculate that from the centers of mass. In real life, every atom is attracted to every other atom in the universe, even minutely, and practically that means the atoms on the side closest to the other object are attracted more strongly than the objects on the opposite side, which means they are effectively being pulled apart (although, usually also very minutely). Let's work with a simple version for now; why the Moon is tidal-locked is an exercise for the interested student.
To make a stable, circular orbit, at all times the pull of gravity must be balanced by something else. Otherwise, the orbiting object will either start accelerating away from the planet/star/whatever, or towards it. The math to derive this get complicated, but luckily the result is simple: to pull a object in a circle, the inwards acceleration is equal to the velocity squared divided by the radius of the curve. In other words, to maintain that particular curve, you must impart a particular acceleration towards the center of the curve. Every action has an equal and opposite reaction, as Isaac Newton said. You feel this when you are in a car going around a curve: the steering wheels are imparting forces that accelerate the car toward the center of the circle the car is turning in. You, loose in the car, feel pushed towards the outside, but that is only because you are in the frame of reference of the car. Objectively, YOU are trying to go in a straight line in accordance with momentum and natural law; the seat and seatbelts (and perhaps the inside of the passenger window if I am driving) are in the way of you going straight while the car turns, and therefore impart a force on you to cause you also to turn. If those forces were not balanced, you would tear through the structure of the car and continue on your straight line until you interacted with other forces, such as the road, a tree, and the young mother on the sidewalk pushing the baby stroller.
We can balance those two equations. In a perfect circular orbit, gravitational acceleration is equal to the acceleration needed to make a circle exactly equal to the distance between the two objects. So:
Circular orbital velocity v = square root ( G * (mass of star) / (orbital radius) )
That orbital velocity isn't just a certain speed in a random direction. It is specifically a velocity normal to the pull of gravity. In 2 dimensions, "normal" means things that are 90 degrees off from each other. In three dimensions, it is all the possible directions that are exactly 90 degrees off. Point towards the center of the earth with your index finger, and stick your thumb out. Rotate around. Every possible way your thumb can point, while your index finger is still pointed to the center of the earth, is 'normal' to the gravitational vector.
The orbital velocity you're going is key: Too fast, and your inwards and outward forces won't match, and you'll accelerate away from the object you want to orbit. Too slow, and you accelerate inward, possible until you have lithobraking (i.e. rapidly slowing down by plowing through rocks).
In the context of the recharge stations, I meant 'free-ride' meaning being in orbit. You don't need to use rockets at all to be in orbit. You fall forever and always miss. But to get that free ride, you have to be going a specific speed. If you're not, you can make up for it using rockets, which means using fuel. In canon, the recharging stations and jumpships are either above or below the star, and they're just hanging out there. Gravity is still there, at about 0.1G. If they don't make 0.1G continuously with rockets, they fall toward the star. They have no orbital velocity, because in canon they need to be at a particular place above the solar poles about 30 AU out. Also, in canon, they lose all momentum when they jump, so immediately after they jump, they fall like a rock towards the star. (Also, they fall like a feather. In vacuum, rocks, feathers and jumpships drop at the same acceleration.) Any version of this where jumpships, recharging stations, etc do not need a continual supply of fresh fuel for their station-keeping nuclear rockets requires them to turn their thrust towards making an free-fall orbit. Luckily, that's achievable in canon; NASA would commit not just murder but genocide for the technology to continuously make 0.1G thrust. (Next problem: running a fusion reactor in space without cooking the crew. Getting rid of heat in a vacuum is HARD.)
This can get more complicated with non-circular orbits and thrust in different directions and complicated even more by having multiple massive objects independently asserting a significant gravitational force on our poor jumpships or dropships (try a BINARY stellar system for true screw-this-I'm-jumping-back-to-Terra times), but hopefully this answers your questions.
Also, just so you know, this was the TL;DR version. This is literal rocket science.
