Start at the dead centre of the Mandelbrot set's big body, c = 0, where the Julia set is a plain circle. Walk out at a 15° heading to the edge, then leave. The whole way, watch one point, z = 0, and where it goes. Inside, it spirals home turning 15° a step. At the edge it takes forever. Outside, it escapes.
Small words. Drop a marble at 0 and keep doing "square it, add c". Inside the big body the marble swirls into a drain, turning a little each time, like water going down a sink. Right at the edge the drain is so weak the marble almost never gets there. Step outside and there's no drain at all, so the marble rolls away forever.
Leg 1, inside the belly (0 → 1): c = λ/2 − λ²/4, with λ = r · e^(i·15°) and r from 0 to 1. The Julia set has one home point that pulls with strength r and turns everything 15° a step. The orbit of 0 spirals into it. checked
The edge (r = 1): the pull is exactly 1, so it barely pulls. 15° is 1/24 of a turn, so this edge point is where a small bulb with period 24 hangs off the belly. The Julia set pinches into 24 petals, and 0 creeps home through them. checked
Leg 2, the exit (1 → 2): leaving along the one outside path that lands on that edge point: external ray 1/(2²⁴ − 1). The walk goes out to height 0.1. From the moment you leave, 0 escapes, and the further out you are, the sooner it goes. checked
▓ Measured turn: over the last 24 steps before it lands, the orbit of 0 turns 15.00° a step at r = 0.9. At the edge it's also 15.00° over 24 steps, though single steps range from 10° to 28° as it hops from petal to petal. checked The page measures this live for wherever you stand. checked
At the edge 0 doesn't really arrive: after 400,000 steps it's still 0.114 from home. checked
The exit path was traced, then read back: 80 points along the ray each report their own height to within 2 parts in a billion. checked Rays land slowly at pinch points: at height 10⁻⁶ the path is still 0.0044 from the edge point. checked
"0 0 0" as c = 0 at phase 0, "tracked by the Julia set at 0 0" as the point z = 0, and "exit trajectory" as the escaping orbit of 0 are my reading of your words. my read
Home point: z* = (1 − √(1 − 4c)) ÷ 2. Its pull is f′(z*) = 2z* = λ. Checked to 10⁻¹² at every r. checked
Edge point (r = 1): c = e^(iπ/12)/2 − e^(iπ/6)/4 = 0.266457 + 0.004410i. Its pull has size 1.000000 at angle 15.000°. checked
15° = 1/24 turn. Two outside paths land on the root of the 1/24 bulb: 1/(2²⁴ − 1) and 2/(2²⁴ − 1). The lower one is used here. checked
On the exit leg, 0 escapes past 2 at step 4 (height 0.1), step 8 (0.01), step 11 (0.001), 14 (10⁻⁴), 18 (10⁻⁵), 21 (10⁻⁶). checked
15° × 24 = 360°: 24 steps of your phase make one full turn, the same 24 as the sentinel cubit. checked