The clock hypothesis of relativity theory equates the proper time experienced by a point particle along a timelike curve with the length of that curve as determined by the metric.
This is indeed the correct general formulation of the clock hypothesis. This formulation applies for all reference frames, inertial or not, and for all spacetimes, flat or curved. The only restriction is that the coordinate basis must have one timelike basis, $dt$.
Because of this, time dilation can be written as $$\frac{1}{\gamma}=\frac{d\tau}{dt}$$ where $\gamma$ is the time dilation factor and $d\tau$ is the proper time, which is related to the metric by $$ds^2=-c^2 d\tau^2=g_{\mu\nu}dx^\mu dx^\nu$$
So, for an inertial frame we have $$ds^2= -c^2 dt^2+ dx^2 + dy^2 + dz^2$$ $$ \frac{d\tau^2}{dt^2}=1-\frac{dx^2}{c^2 dt^2}-\frac{dy^2}{c^2 dt^2}-\frac{dz^2}{c^2 dt^2}$$ $$\frac{1}{\gamma}=\sqrt{1-\frac{v^2}{c^2}}$$ which is the usual familiar time dilation formula.
Note that it has the property that it depends only on the velocity and not the acceleration. When calculating $d\tau/dt$ we get terms like $dx^2/dt^2$, which is the square of a component of velocity, not an acceleration like $d^2x/dt^2$ would be.
Therefore, unlike the other version of the postulate, this one works in non-inertial frames and seems to be consistent with acceleration directly causing time dilation
Although the general formula does work in non-inertial frames, it still does not give a situation where acceleration directly causes time dilation. Let’s work it out for the Rindler metric you mentioned. For convenience I will use units where $c=1$: $$ds^2= -(gx)^2 dt^2+ dx^2+dy^2+dz^2$$ $$\frac{d\tau^2}{dt^2}=(gx)^2-\frac{dx^2}{dt^2}-\frac{dy^2}{dt^2}-\frac{dz^2}{dt^2}$$ $$\frac{1}{\gamma}=\sqrt{(gx)^2-v^2}$$
Now, you might be tempted to say “look, it has $g$ which is a pseudo gravitational acceleration, so the Rindler time dilation is directly related to acceleration”. However, closer inspection shows that it is not just $g$, but $gx$ which is the form of a gravitational potential, not gravitational acceleration.
Furthermore although $g$ has units of acceleration, it is a property of the particular coordinates chosen, not the acceleration of any worldline. The time dilation of a specific object does not depend on that object’s acceleration, only its position and velocity with respect to the chosen coordinates. If an object is momentarily at rest ($v=0$) at some $x=x_0$ then it’s time dilation is $\gamma=1/(gx_0)$, regardless of what that object’s acceleration is. So this time dilation term is a function of position, not acceleration.
It turns out that this is a general fact. In the metric the terms $g_{\mu\nu}$ are functions of position only, not velocity nor acceleration. And so when you divide by $dt^2$ you only get terms that are functions of position times velocities, like $$g_{xy}\frac{dx}{dt}\frac{dy}{dt}$$ You simply cannot get anything else because of the form of the metric.
Best Answer
Yes. This is basically the triangle inequality. In Euclidean geometry lengths are given by $ds^2 = dx^2 + dy^2 + dz^2$. This metric results in the fact that the shortest path between two points is a straight line.
In Minkowski (or Lorentzian) geometry, lengths are given by $ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2$. This metric results in the fact that the longest path between two points is a straight line. At least for the paths that massive objects can take.
So the elapsed time on a watch, called proper time, is just the length of your path through spacetime. That length is longest for a straight path because of the - sign in the metric.