Function of the Accelerometer
The accelerometer measures the time derivative of its own momentum as it would be measured by an inertial frame with the same velocity. However, it does not display the measured value. It displays the measured value divided by its pre-programmed known rest mass.
$$\frac{dp}{dt} := F = \frac{d(mv)}{dt} = \frac{dm}{dt}v +\frac{dv}{dt}m$$
For low relative velocities between the accelerometer and the measurement frame W, $m = m_0$, $\frac{dm}{dt} = 0$ and we can simplify to $F = m_0\frac{dv}{dt}$. The accelerometer divides by the pre-programmed known rest mass and displays $\frac{dv}{dt}$.
Predicting the Accelerometer's Display from an Arbitrary Inertial Frame
For an object A rotating around an accelerating center of rotation C at radius r, angular velocity $\omega$, all as measured by W: $$\frac{d\vec v}{dt} = \vec a_A = \vec a_C - r\omega^2 \hat r$$
If we can make the approximation $m = m_0$, we're done: when W looks through her telescope at the accelerometer, that's the value she will read.
If you want to extend the result to relativistic relative velocities (if we run the experiment for a long time, such that v as measured by W is a significant fraction of c):
$$m=\gamma(v) m_0$$ where $$\gamma(v) = \frac{1}{\sqrt{1-v^2/c^2}}$$
then carrying out the time derivative of momentum gets:
$$\frac{d\vec p}{dt} = \frac{\gamma^3 m_0}{c^2}(\vec v \cdot \vec a)\vec v + \gamma m_0 \vec a$$
and W can predict that if W looks through a telescope at the accelerometer display it will read
$$\frac{\gamma^3}{c^2}(\vec v \cdot \vec a)\vec v + \gamma \vec a$$
where $\vec a$ is as calculated above. Note that for small relative velocities ($v<<c$), $\gamma \to 1$ so the accelerometer displays $\vec a$ as you would expect.