For an observer who is not in a gravitational field with any relevant strength, the coordinate speed of light in a remote point under the influence of stronger gravitation is not the universial constant $c_0$ but some lower value.
But what is the coordinate speed of light? Huuh, let me try an explanation, but with the disclaimer that I am not fully sure about it. Consider the observer cutting a stick from a tree and calls its length a quack. He then defines that the time light takes from one end of the stick to the other and back is called a quick. So distance and time are defined and the speed of light is $1\frac{\text{quack}}{\text{quick}}$.
He now moves to the remote point in question. Trivially, the local speed of light is again $1\,\text{quack}/\text{quick}$. But if he compares how long a beam of light takes along the stick and back with a same-length stick left behind at his home location (and doing this comparison is not at all trivial), he will find it takes longer locally, so the speed of light measured with his coordinates brought along from home is lower. That's why it is coordinate speed of light. In contrast, the local speed of light is pretty boring, as it trivally is always the same $1\text{quack}/\text{quick}$.
What is the coordinate speed of light, given the metric? For the Schwarzschild Metric I got a nice answer on Physics StackExchange.
Basically the same derivation can be done for a general metric $ds^2 = g_{\mu\nu} dx^\mu dx^\nu$. The coordinate system we use is set up so that the 0-index coordinate goes in the direction of time, i.e. $dx^0 = c_0 dt$? Divide by $dt^2$ to get $$ -g_{00} c_0^2 = \frac{g_{ij}dx^i dx^j}{dt^2} + 2 c_0 \sum_{i=1}^3 \frac{g_{0i} dx^i}{dt} \qquad i,j\in\{1,2,3\} $$ using the fact that the metric is symmetric, hence the simple sum times $2$. Also note the use of the Einstein summation also for when the indices only run over $\{1, 2, 3\}$.
Now comes the same trick as in the StackExchange answer cited above. The observed velocity $c$ in an arbitrary direction is $c \vec{e}$ where $|\vec{e}| = |(e_1, e_2, e_3)| = 1$ is a unit vector and $dx^i/dt = c e_i$. So we somewhat trivially get: $$ -g_{00} c_0^2= g_{ij}e^i e^j c^2 + 2 c_0 c \sum_{i=1}^3 g_{0i} e^i \qquad i,j\in\{1,2,3\} $$ With $a = g_{ij}e^i e^j$ and $b = \sum_{i=1}^3 g_{0i} e^i$ this is \begin{align*} -g_{00} c_0^2 &= a c^2 + 2 c_0 c b \\ -g_{00} c_0^2 /a &= c^2 + 2 c_0 c b/a \\ -g_{00} c_0^2 /a + (c_0 b/a)^2 &= (c_0 b/a + c)^2 \\ c &= \sqrt{ -g_{00} c_0^2 /a + (c_0 b/a)^2 } - c_0 b/a \\ &= c_0 \sqrt{ -g_{00} /a + (b/a)^2 } - c_0 b/a \\ &= \frac{1}{a} \left( \sqrt{ b^2 -g_{00} } - b\right) c_0 \\ &=: \alpha(\vec{e}) c_0 \end{align*} Dragging along the $c_0$ and the $c$ instead of just working with $c_0=1$ allows a simple unit check, and yes, both sides are scalar velocities.
Another quick test: In the Minkowski Metric, the matrix with the diagonal $(1, -1, -1, -1)$, we get $a =g_{ij}e^ie^j = -1 \sum_{j=1}^{3} (e^j)^2 = -1$ and $b=0$, so we have $c = c_0 \sqrt{-1 / -1} =c_0$, as required.
The metric tensors coefficients $g_{\mu\nu}$ are not invariant with regards to Lorentz transformations of the coordinate system. The particular coefficients have no physical meaning. But what about $\alpha(\vec{e})c_0$, the speed of light in the direction of $\vec{e}$ at the point we are looking at. It should clearly not depend on our choice of coordinate system. And so $\alpha(\vec{e})$ shouldn't. Lets check.
Under a Lorentz transformation the product $g(v,w) := v^T g w$ for a given metric $g$ and two vectors $v$ and $w$, is invariant. Below we show that each of the terms of $\alpha$ are of the form $g(v,w)$ and therefore invariant.
In the above, the direction vector $\vec{e}$ was explicitly described as three-dimensional. But from now on, lets define it as four dimensional as $\vec{e} = (0, e^1, e^2, e^3)^T$, where the coefficients are the ones from the three dimensional vector above. Lets also have $\vec{t}:=(1, 0, 0, 0)^T$. Then each term of $\alpha$ is some combination of $g$, $\vec{t}$ and $\vec{e}$.
Since all terms appearing in the definition of $\alpha$ are invariant, we have shown that $\alpha$ itself is invariant as required for sanity.
Like others on PSE I wondered what the coefficients of $g$ tell us about the point in space where they apply. It was frustrating to understand that the answer is: nothing, beyond angles and lengths, but dependent on the coordinate system. Change your coordinate basis and the values can be all different.
In contrast $\alpha(\vec{e})$ is invariant, i.e. independent on the choice of coordinate system.
Do I even know what I am talking about here? The local coordinate system is also a coordinate system and there the (local) speed of light is $c_0$. Now what?It is the local speed of light at a given point in space as measured in a somewhat "sane" coordinate system of an observer at rest with regard to the point and not under the influence of gravity. It is a physical quantity independent of the coordinate system with a rather intuitive meaning. The prototypical example being the speed of light near a large mass, typically somewhat outside the event horizon of a black hole.
But forget black holes. Earth has mass too. And the above tells us the speed of light in one and the same point near a mass can be different, depending on the direction light is traveling. For example, solving the equation in my question on PSE for radial and tangential speed of light, we get. \begin{align} \frac{dr}{dt} &= c_0 \left(1-\frac{r_s}{r} \right)\\ r^2\frac{d\Omega}{dt} &= c_0\sqrt{1-\frac{r_s}{r}} \end{align} In principle we should be able to measure this. Build two identical light clocks let one point upward for the radial direction and fix the other horizontally for the tangential direction. The vertical one should run slower, since $1-r_s/r$ is smaller than its square root.
...at least for me: What can we make of it? For example: