In order to better explain all this to myself (and hopefully to others as well), I will present a simple toy model. Consider two point particles with positions ##q_1(t)## and ##q_2(t)## moving in one dimension. Suppose that the Lagrangian has the form
$$L = \frac{m_1\dot{q}_1^2}{2} + \frac{m_2\dot{q}_2^2}{2} - V_1^{\rm ext}(q_1) - V_2^{\rm ext}(q_2) - \frac{k(q_1-q_2)^2}{2} $$
The external potentials ##V_1^{\rm ext}## and ##V_2^{\rm ext}## describe external forces on the particles, e.g., the Newton gravitational force, the Coulomb electric force, etc. The last term describes the internal force between the particles, which models the working of an accelerometer made of a spring attached to a moving object. The equations of motion are
$$m_1 \ddot{q_1} = F_1^{\rm ext}(q_1) -k (q_1-q_2)$$
$$m_2 \ddot{q_2} = F_2^{\rm ext}(q_2) -k (q_2-q_1)$$
Now consider a special case in which ##F_2^{\rm ext}(q_2) \equiv 0##, meaning that the external force acts only on the first particle. An example is the electric force, for the case in which the first particle is charged while the second particle is neutral. In this case the second equation of motion can be written as
$$\ddot{q_2} = -\frac{k}{m_2} (q_2-q_1)$$
so by observing the relative position ##q_2-q_1## one can determine the acceleration ##\ddot{q_2}##. That's how the accelerometer measures the acceleration.
What if ##F_2^{\rm ext}(q_2)## is not zero? In that case there is no simple relation between the particle positions and the acceleration, so the accelerometer cannot work in a way we are used to. In principle the acceleration still can be determined, but it's much more complicated and depends on details of the external force.
The above suggests that perhaps, in some complicated way, the accelerometer could measure even the gravitational force. To see if this is true, let us explore that case in more detail. For the Newton gravitational force we have
$$ F_i^{\rm ext}(q_i) = m_i g(q_i)$$
where ##g(x)=-\partial_x\phi(x)## and ##\phi(x)## is the external gravitational potential. The equations of motion are then
$$m_1 \ddot{q_1} = m_1g(q_1) -k (q_1-q_2)$$
$$m_2 \ddot{q_2} = m_2g(q_2) -k (q_2-q_1)$$
or equivalently
$$m_1 (a_1 - g(q_1))= -k (q_1-q_2)$$
$$m_2 (a_2 - g(q_2)) = -k (q_2-q_1)$$
where ##a_i \equiv \ddot{q_i}##. Summing the equations we get
$$m_1 (a_1 - g(q_1)) + m_2 (a_2 -g(q_2)) =0$$
But ##m_1## and ##m_2## are both positive, and we assume that the two particles are not too far from each other so that we can use the homogeneous field approximation ##g(q_1)= g(q_2)##. Hence, assuming also the equilibrium ##a_1=a_2##, the equation above implies
$$a_1 - g(q_1)=a_2 -g(q_2)=0$$
so the equations of motion imply
$$q_1-q_2=0$$
Thus the accelerometer always shows a zero reading, independently on the acceleration ##a_1=a_2##. That is the dynamical reason why the gravitational force cannot be determined by the accelerometer.