One of the best ways to describe the situation is to use four-vectors. On the minus side it's a bit technical, on the plus side ambiguities of language don't affect the answers.
Using the 4-vector approach, geometric units, and setting the constant acceleration to unity, the results from for instance http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html would appear as below.
Note that I'm only writing out two of the components of the 4-vector, the other two components are identically zero.
\tau = proper time
The position vector can be written as a function of proper time as follows:
x^i(\tau) :
t = x^0 = sinh \, \tau \hspace{.5 in} d = x^1 = cosh \, \tau
I've chosen this as the simplest way of writing down the equations of motion, note however that x^1(0) is not equal to zero, but is rather equal to 1.
The 4-velocity vector can be written as a function of proper time as follows:
v^0 = \frac{d x^0}{d \tau} = cosh \, \tau \hspace{.5 in} v^1 = \frac{d x^1}{d\tau} = sinh \, \tau
The energy-momentum 4-vector is just
p^0 = m v^0 = m \, cosh \, \tau \hspace{.5 in} p^1 = m v^1 = m \, sinh \, \tau
Here m is the invariant mass of the dime. The simplest case to analyze is where the invariant mass m is constant. This requires the dime to be propelled by an external source, for instance the 'dime' might really be a laser powered light sail.
Note that p^0 can be interpreted as the energy of the dime (it's written here as a function of proper time), and p^1 can be interpreted as the momentum of the dime.
Note that the 4-velocity of the dime is the rate of change of the position of the dime with regard to proper time tau, rather than the coordinate time t.
The 4-force on the dime is
f^0 = \frac{d p^0}{d \tau} = m \, sinh \, \tau \hspace{.5 in} f^1 = \frac{d p^1}{d \tau} = m \, cosh \, \tau
The 4-acceleration of the dime is
a^0 = \frac{d v^0}{d \tau} = sinh \, \tau \hspace{.5 in} a^1 = \frac{d v^1}{d \tau} = cosh \, \tau
Some comments here are in order:
0) There's no problem in transforming the 4-vectors into an instantaneous comoving frame - we can transform both the 4-acceleration and the 4-force in this manner, because 4-vector transformations depend only on the relative velocity, i.e. all 4-vectors transform according to the Lorentz transform, which is indepenent of acceleration and depends only on velocity.
1) using the 4-vector approach, we can say that that the 4-force is directly proportional to the 4-acceleration, and the constant of proportionality is the invariant mass m of the dime.
2) In the frame of the dime, the 4-acceleration is constant. If you do the appropriate Lorentz boost, the 4-acceleration in an instantaneously comoving frame is just (0,1).
3) In the frame of the dime, the 4-force is also constant, and equal to m*a, and since a=1, the 4-force is equal to m.
5) We could write the 4-velocity and energy -momentum 4-vector in the instantaneous frame of the dime just as we did for the 4-acceleration and the 4-force, but they aren't very interesting.
6) The idea of constant proper acceleration is the same as the idea that the 4-acceleration in the instantaneously comoving frame is just (0,a). I've made a=1 for simplicity in writing the above expressions down.
