= Solution
The age comparison is a local comparison of <clocks> at two shared <spacetime events>: departure and reunion. No convention about the simultaneity of distant events is needed to read two adjacent <clocks>. The result is determined by the <proper time> along each complete <world line>. For an ideal <clock> following a <timelike curve> in the home observer's <inertial frame>,
$$
\Delta\tau=\int_0^T\sqrt{1-\frac{\dot x(t)^2}{c^2}}\,dt.
$$
The home observer stays at $x=0$, whereas the traveler goes outward at <velocity> $v$ and returns at <velocity> $-v$. Ignoring the duration of a short turnaround gives home age $T$ and traveler age $T/\gamma$. The <twin paradox with fixed reunion events> derives this directly. There is no contradiction because the traveler does not remain in one <inertial frame> throughout the trip.
<Relativity of simultaneity> becomes essential to a different question: how can the traveler consistently describe the home <clock> as running slowly during both straight legs? The statement about a distant <clock>'s rate refers to a sequence of remote events selected as simultaneous by a specified <inertial frame>. Changing that frame changes the selected sequence. Adding the two slow-rate comparisons while silently preserving one distant-time assignment is the error.
Let turnaround be $P=(t,x)=(T/2,vT/2)$ in the home frame. Constant outward-frame time means $t-vx/c^2$ is constant. Thus the home event simultaneous with $P$ according to the outward frame occurs at
$$
t_H^-=\frac T2\left(1-\frac{v^2}{c^2}\right).
$$
For the return frame, constant time means $t+vx/c^2$ is constant, so its simultaneous home event occurs at
$$
t_H^+=\frac T2\left(1+\frac{v^2}{c^2}\right).
$$
These are two different remote <spacetime events>, even though both assignments pass through the same turnaround event. The <twin paradox simultaneity change> is
$$
t_H^+-t_H^-=\frac{v^2T}{c^2}.
$$
Each leg contributes $T/(2\gamma^2)$ to the home <clock> advance assigned within that leg's frame. Including the reassignment gives $T/\gamma^2+v^2T/c^2=T$, exactly the home reading at reunion. For the illustrative values $v=0.6c$ and $T=10$ years, the traveler's elapsed <proper time> is eight years; the two leg assignments give $3.2+3.2$ years of home advance, and the frame change supplies the missing $3.6$ years. The distant home <clock> does not physically jump.
Nor should simultaneity assignments be confused with received <light> signals. An observer sees emissions after a propagation delay and measures a <Doppler shift>. The received history remains consistent through a finite turnaround, but it is not the same sequence of events as a constant-coordinate-time slice. The <Lorentz transformation> relates the latter; signal travel times must be included separately when discussing what is seen.
An instantaneous turnaround is only an ideal limit. In the <twin paradox with finite acceleration>, the <world line> and the momentarily comoving frame vary continuously. The distant home-time assignment is $t_H(t)=t-v(t)x(t)/c^2$, so
$$
\frac{dt_H}{dt}=1-\frac{v(t)^2}{c^2}-\frac{x(t)}{c^2}\frac{dv}{dt}.
$$
The reversal at positive $x$ makes the last contribution positive. Its integrated effect replaces the ideal reassignment. By the <clock hypothesis>, the traveler's own elapsed <proper time> has no extra term depending directly on <acceleration>; physical disturbances of a particular <clock> are a separate experimental issue.
Thus <relativity of simultaneity> explains the consistency of the traveler's successive inertial descriptions. It is not a convention-dependent cause of the final age difference. The <invariance of reunion ages under clock resynchronization> makes that distinction explicit: remote time labels can change while the two reunited <clock> readings remain unchanged.
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