What I understand so far
Fix a point $A$ with coordinates $x^\alpha$ with respect to a frame $K$ with metric tensor $g_{\alpha\beta}$. Then take an arbitrary point $B$ infinitesimally close to it with coordinates $x^\alpha+dx^\alpha$. Suppose that a light signal is directed from $B$ to $A$ and then back over the same path in the frame of $B$, which we shall denote with $K'$. Let $\tau$ be proper time in the $K'$ frame and $d\tau$ be the change in time between emission of the signal and return of it, both of which occur at the same spatial coordinates in both $K$ and $K'$. From this, we would deduce the formula
$$d\tau = \frac{1}{c}\sqrt{g_{00}}dx^0. \qquad (1)$$
Then, in $K'$, letting $dl$ be the infinitesimal distance between $A$ and $B$, we have
$$\begin{eqnarray}0 = ds^2 &=& c^2 \left(\frac{d\tau}{2}\right)^2 - dl^2, \\ &=& g_{00}\left(\frac{dx^0}{2}\right)^2 - dl^2. \end{eqnarray} \qquad (2) $$
where $ds$ is the spacetime distance between $A$ and $B$ and $\tau$ is proper time or the time of clock at $B$ and $d \tau$ is the time between light signal emission from $B$ and its reception at $A$. Here, we have done an Einstein summation over the space indices $1,2,3$ and used the fact that light-like intervals have spacetime distance of $0$.
By the invariance of the spacetime interval, we also have, per the metric tensor,
$$0 = ds^2 = g_{\alpha\beta} dx^\alpha dx^\beta + 2g_{0\alpha} dx^0 dx^\alpha + g_{00}(dx^0)^2, \qquad (3)$$
which is a quadratic in $dx^0$ and thus has two solutions
$$ \begin{eqnarray} dx^{0(1)} &=& \frac{1}{g_{00}}\{-g_{0\alpha}dx^\alpha - \sqrt{(g_{0\alpha}g_{0\beta} - g_{\alpha\beta}g_{00})dx^\alpha dx^\beta}\},\\ dx^{0(2)} &=& \frac{1}{g_{00}}\{-g_{0\alpha}dx^\alpha + \sqrt{(g_{0\alpha}g_{0\beta} - g_{\alpha\beta}g_{00})dx^\alpha dx^\beta}\}. \end{eqnarray} \qquad (4) $$
How to interpret these? We notice that for the $dx^\alpha$ to not flip sign, they must in both cases correspond to spatial coordinates of point $B$ minus spatial coordinates of point $A$. With this in mind, the negative solution $dx^{0(1)}$ corresponds to, in frame $K$, the time of emission from $B$ minus the time of reception at $A$, and positive solution $dx^{0(2)}$ corresponds to time of return of signal to $B$ minus time of reception at $A$.
Thus, the $dx^0$ in $(1), (2)$, which is the time elapsed in $K$ through the entire light signal exchange process, corresponds to
$$dx^0 = dx^{0(2)}-dx^{0(1)} = \frac{2}{g_{00}}\sqrt{(g_{0\alpha}g_{0\beta} - g_{\alpha\beta}g_{00})dx^\alpha dx^\beta}, \qquad (5)$$
which one can substitute into $(2)$ to solve for $dl$.
What I don't understand
I'll copy directly from Landau-Lifshitz Classical Theory of Fields, and write in italics my questions or comments.
We now turn to the definition of the concept of simultaneity in the general theory of relativity. In other words, we discuss the question of the possibility of synchronizing clocks located at different points in space, i.e. the setting up of a correspondence between the readings of these clocks. (As for simultaneity of relativity, I have not studied it in detail, but I'm well aware that whether or not two events happen at the same time depends on the reference frame. My understanding of clock synchronization in relativity is also quite superficial. However, I am aware that the synchronization condition is that when point B sends light to a mirror at point A which returns, A's clock at the time when it receives the light should be the average of the emission and return times of B's clock. Of course, this is also a rather vague, trivial statement of consistency between clocks at two points, and I do not quite grasp the significance of this to the bigger picture. I welcome knowledgeable people to recommend a good source on this.)
Such a synchronization must obviously be achieved by means of an exchange of light signals between the two points. We again consider the process of propagation of signals between two infinitely near points A and B. We should regard as simultaneous with the moment $x^0$ at the point $A$ that reading of the clock at point $B$ which is half-way between the moments of departure and the return of the signal to that point, i.e. the moment
$$x^0 + \Delta x^0 = x^0+\frac{1}{2}(dx^{0(2)}+dx^{0(1)}). \qquad (6)$$
The above equation clearly represents the midpoint of $x^0+dx^{0(1)}$ and $x^0+dx^{0(2)}$, which are the emission and return times in the $K$ frame. As for the clock at $B$ or the $K'$ frame, we know its rate of passage relative to the rate of passage of the clock at $K$ during the signal exchange process to be given by $(1)$. We also know that in that frame, the arrival time of the light signal at $A$ must be the midpoint of the emission time and return time. On the other hand, if $(3)$ has two solutions, then in the $K$ frame, the time of arrival at $A$ is not the midpoint of emission time and return time, and $x^0$ in $(6)$ represents deviation from that midpoint.
Substituting $(4)$, we thus find that the difference in the values of the "time" $x^0$ for two simultaneous events occurring at infinitely near points (The algebraic manipulation here is straightforward, but I am unclear as to what is actually represented physically. What exactly is this "time" $x^0$? What exactly are these "two simultaneous events"? As far I can tell, one is the arrival of light signal at $A$ and the other is the reading of the clock at $B$ at that event, the two of which are simultaneous because they were both directly with respect to the same physical event, with one of them at a different spatial location of the event itself.) is given by
$$\Delta x^0 = -\frac{g_{0\alpha}dx^\alpha}{g_{00}} \equiv g_{\alpha}dx^{\alpha}. \qquad (84.14)$$
This relation enables us to synchronize clocks in any infinitesimal region of space. (How, in detail? What does it mean for clocks to be synchronized in a region of space?) Carrying out a similar synchronization from the point $A$, we can synchronize clocks, i.e. we can define simultaneity of events, along any open curve. $\dagger$ (An ideal answer would explain this paragraph in full detail, with no assumptions of prior knowledge.)
However, synchronization of clocks along a closed contour turns out to be impossible in general. In fact, starting out along the contour and returning to the initial point, we would obtain for $\Delta x^0$ a value different from zero. (Please give full detail, formalism, and proof for this.)
$\dagger$ Multiplying $(84.14)$ by $g_{00}$ and bringing both terms to one side, we can state the condition for synchronization in the form $dx_0 = g_{0i}dx^i = 0$ (At least superficially, this appears to be enforcing $\Delta x^0 = 0$? Is that case, and if so, why would such be reasonable?): the "covariant differential" $dx_0$ between two infinitely close simultaneous events must be zero.
