Consider ordinary Minkowski spacetime. In standard cartesian coordinates $(t,x^1,x^2,x^3)$ - in which the speed of light is isotropic - the line element takes the form
$$\mathrm ds^2 = -c^2 \mathrm dt^2 + \sum_{i=1}^3 (\mathrm dx^i)^2\tag{1}$$
In these coordinates, the fact that a light ray travels along a null worldline implies that along that worldline, $$\mathrm ds^2 = 0 \implies \sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2 = c^2$$
Now we choose other coordinates $(T,x^1,x^2,x^3)$, where the new time coordinate is $T = t - x/c$. The line element now takes the form
$$\mathrm ds^2 = -c^2 \mathrm dT^2 + \sum_{i=2}^3(\mathrm dx^i)^2 - 2c\big( \mathrm dT \mathrm dx^1\big)\tag{2}$$
This is the same spacetime and the same metric - just an unconventional choice of coordinates. Furthermore, the null condition now takes a different form. Restricting our attention to motion along the $x^1$ direction, the null condition becomes $-c^2 \mathrm dT^2- 2c\big(\mathrm dT \mathrm dx^1\big) = 0$ If $\mathrm dx^1>0$ (so the ray is moving to the right) then the only possibility is that $\mathrm dT =0$ (so the velocity $\mathrm dx^1/\mathrm dT$ is formally infinite). On the other hand, if the ray is moving to the left then for future-directed null worldlines we have that $\mathrm dx^1/\mathrm dT = -c/2$.
As you can see, whether the velocity (which is a coordinate-dependent quantity, after all) is isotropic or not depends entirely on our choice of time coordinate. When one adopts the Einstein summation convention, they obtain $(1)$, but this is not mandatory.
Presumably, their two-way speed limit would be $c$, but if the one-way limit changes with direction, wouldn't they be speeding up and slowing down (e.g., between $2/3 c$ and $2c$) as they move around the loop? If so, what forces would be acting on them to cause such changes? Wouldn't they feel these forces and notice that things appear to be moving by them at changing speeds?
No forces would be necessary, and nothing out of the ordinary would be felt. If you run through a clock shop and define your velocity as the distance between adjacent clocks divided by the difference in their readings, then you could be moving at what you consider a stationary pace but the numerical value of your velocity would change if the clocks are not synchronized across the shop. This is essentially the same concept - velocity is not measurable in a coordinate-independent way, a central point of special relativity.
Concretely, consider the worldline $(t,x,y)=\big(\lambda,R \cos(\omega \lambda), R\sin(\omega \lambda)\big)$ in coordinate system $(1)$, corresponding to a particle moving in a circle. Its easy to verify that the speed of the particle in these coordinates is constant, and equal to $\omega r$.
In our new coordinates $(2)$, this becomes
$$(T,x,y)=\big(\lambda-\frac{r}{c}\cos(\omega\lambda),r\cos(\omega\lambda),r\sin(\omega\lambda)\big)$$
$$\implies \frac{dx}{dT} = -\frac{\omega r \sin(\omega\lambda)}{1+\frac{\omega r}{c}\sin(\omega\lambda)} \qquad \frac{dy}{dT} = \frac{\omega r \cos(\omega\lambda)}{1+\frac{\omega r}{c}\sin(\omega\lambda)}$$
$$\implies \sqrt{\left(\frac{dx}{dT}\right)^2 + \left(\frac{dy}{dT}\right)^2} = \frac{\omega r}{1+\frac{\omega r}{c}\sin(\omega\lambda)} = \frac{\omega r}{1+\frac{\omega y}{c}}$$
Therefore, the speed as calculated in these coordinates is not constant.
It should be obvious that nothing has physically changed here - we're just using new coordinates. The trajectory of the particles as observed by a human looking at it with their eyes in a laboratory is unchanged. However, speed is by definition a coordinate dependent notion, and using unconventional coordinates yields to unconventional results like this.
In this case, is our speed limit still $c$ or is it $2/3 c$, $2c$, or something else? Does it depend on the direction of travel?
The "speed limit" is a manifestation of the condition that the worldline of a massive particle must be timelike. In coordinate system $(1)$, this means that $\sqrt{\sum_{i=1}^3 \left(\frac{dx^i}{dt}\right)^2} < c$, making the term speed limit a reasonable one.
In coordinate system $(2)$, the same condition means that
$$-2c\big(\mathrm dT \mathrm dx^1) + \sum_{i=2}^3 \big(\mathrm dx^i\big)^2 < c^2 \mathrm dT$$
This is harder to cast into the form of a speed limit in general. However, if $\mathrm dx^1=0$ so the motion is occuring in the $(x^2,x^3)$-plane, then we obtain the same limit as before, while if the motion is occurring along the $x^1$ axis only then $\mathrm dx^1/\mathrm dT \in (-c/2, \infty)$.
Best Answer
The Einstein synchronization convention produces a one-way speed of light that is c. So the second postulate is based on the one way speed. This is justified by the isotropy of the two way speed of light and the isotropy of all known laws of physics.
In Einstein’s seminal paper he says “we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A.” Where those two times are the one-way times and setting them equal makes the assumption that the one way speed equals the two way speed.