If you want to know how old a traveling twin will be when he gets to his destination, you can always use the formula:
[itex]\Delta \tau = \int_{t_A}^{t_B} \sqrt{1-\frac{v^2}{c^2}} dt[/itex]
where [itex]t_A[/itex] is the time at the start of the journey, [itex]t_B[/itex] is the time at the end of the journey, and [itex]v[/itex] is the speed of the traveler (which might be changing---it doesn't matter). You can pick any inertial coordinate system to measure [itex]t_A, t_B, v[/itex] and [itex]dt[/itex], and you will get the same answer. The critical point is the word "inertial". If the one twin turns around and returns, then there is no inertial reference frame in which he is always at rest.
You can use a noninertial coordinate system, also, but the formula has to be modified:
[itex]\Delta \tau = \int \sqrt{\sum_{\alpha \beta} g_{\alpha \beta} \frac{dx^\alpha}{ds} \frac{dx^\beta}{ds}} ds[/itex]
where [itex]g_{\alpha \beta}[/itex] is the components of the metric tensor in that coordinate system, and [itex]x^\alpha[/itex] are the coordinates in that reference frame, and [itex]s[/itex] is any parametrization of the journey (it could be proper time, or it could be anything that increases smoothly along the journey).