The refer to the simplified answer for the transformation between the inertial frame unprimed and the accelerated primed frame:

t = gt' + g(v/c²)[x'cos(wgt') + y'sin(wgt')]

x = x'[gcos²(wgt') + sin²(wgt')] + r₀sin(wgt') + (g - 1)y'sin(wgt')cos(wgt')

y = y'[cos²(wgt') + gsin²(wgt')] - r₀cos(wgt') + (g - 1)x'sin(wgt')cos(wgt')

The counter rotating frame transformation to the inertial frame will also be

t = gt" - g(v/c²)[x"cos(wgt") - y"sin(wgt")]

x = x"[gcos²(wgt") + sin²(wgt")] - r₀sin(wgt") - (g - 1)y"sin(wgt")cos(wgt")

y = y"[cos²(wgt") + gsin²(wgt")] - r₀cos(wgt") - (g - 1)x"sin(wgt")cos(wgt")

From these relate the two accelerated frame coordinates

gt' + g(v/c²)[x'cos(wgt') + y'sin(wgt')] = gt" - g(v/c²)[x"cos(wgt") - y"sin(wgt")]

x'[gcos²(wgt') + sin²(wgt')] + r₀sin(wgt') + (g - 1)y'sin(wgt')cos(wgt') = x"[gcos²(wgt") + sin²(wgt")] - r₀sin(wgt") - (g - 1)y"sin(wgt")cos(wgt")

y'[cos²(wgt') + gsin²(wgt')] - r₀cos(wgt') + (g - 1)x'sin(wgt')cos(wgt') = y"[cos²(wgt") + gsin²(wgt")] - r₀cos(wgt") - (g - 1)x"sin(wgt")cos(wgt")

Consider what the double primed frame observer's watch reads according to the single primed frame. It is at the origin of the double primed frame, so:

gt' + g(v/c²)[x'cos(wgt') + y'sin(wgt')] = gt"

x'[gcos²(wgt') + sin²(wgt')] + (g - 1)y'sin(wgt')cos(wgt') = - r₀[sin(wgt") + sin(wgt')]

y'[cos²(wgt') + gsin²(wgt')] + (g - 1)x'sin(wgt')cos(wgt') = r₀[cos(wgt') - cos(wgt")]

Solve the last equation for y' and insert into the other two and simplify.

y'= {r₀[cos(wgt') - cos(wgt")] - (g - 1)x'sin(wgt')cos(wgt')}/ [cos²(wgt') + gsin²(wgt')]

gt'[1 +(g - 1)sin²(wgt')] + g(v/c²)x'cos(wgt') + g(v/c²)sin(wgt')r₀[cos(wgt') - cos(wgt")] = gt"[1 +(g - 1)sin²(wgt')]

x'g = - r₀sin(wgt")[1 + (g - 1)sin²(wgt')] - gsin(wgt')r₀ + (g - 1)sin(wgt')cos(wgt')r₀cos(wgt")

Substitute the last into the previous equation and simplify

gt' - r₀(v/c²)sin[wg(t" + t')] - gt" = 0

This says that the watch re-synchs with the single primed frame observer's watch at

wgt' = wgt" = np/2

These include the locations where the rockets meet.

Differentiate

dt' - w(dt" + dt')r₀(v/c²)cos[wg(t" + t')] - dt" = 0

Consider those re-synch points where they meet at wgt' = wgt" = mp

dt' - w(dt" + dt')r₀(v/c²)cos(2mp) - dt" = 0

dt' - w(dt" + dt')r₀(v/c²)cos(2mp) - dt" = 0

Simplify to

dt' = (1 + v²/c²)dt"/(1 - v²/c²)

Let u = 2v/(1 + v²/c²)

Inserting this and simplification results in

dt' = dt'/(1 - u²/c²)^1/2

This demonstrates that one rocket frame observer will observe the other as time dilated by special relativistic time dilation as they pass each other.