Relative velocity and proper time derivative of geodesic deviation? From wiki

To quantify geodesic deviation, one begins by setting up a family of closely spaced geodesics indexed by a continuous variable s and parametrized by an affine parameter $\tau$. That is, for each fixed $s$, the curve swept out by $\gamma _s(\tau)$ as τ varies is a geodesic. When considering the geodesic of a massive object, it is often convenient to choose $\tau$ to be the object's proper time. If $x^μ(s, τ)$ are the coordinates of the geodesic $\gamma_s(\tau)$, then the tangent vector of this geodesic is:
$$ T^\mu = \frac{\partial x^\mu(s,\tau)}{\partial \tau} $$
If $\tau$ is the proper time, then $T_\mu$ is the four-velocity of the object traveling along the geodesic.
One can also define a deviation vector, which is the displacement of two objects traveling along two infinitesimally separated geodesics:
$$ X^\mu =  \frac{\partial x^\mu(s,\tau)}{\partial s} $$

Now, naively
$$ v^\mu = T^\beta \nabla_\beta X^\mu$$
Should $v^\mu$ be the relative velocity? If so, how does that fit in with this definition of relative velocity? (when their geodesics intersect)

$$
v:=-\frac{1}{g(u^′,u)} u^′ - u \tag{1}
$$

I feel I've managed to confuse myself.
 A: Indeed, you seem to have confused yourself :-)
I edited several of your formulas. Check them out.
The equation
$$ v^\mu = T^\beta \nabla_\beta X^\mu$$
measures the rate of separation between 2 neighboring geodesics, so it measures their relative 4-velocity.
The quoted question and the paper mentioned therein take the matter to a deeper level, however...
you have to realize that the "relative ordinary velocity" of an object passing close to you, regardless if it is moving on a geodesic or not, is simply the spacelike component of its 4-velocity (corrected by a $\gamma$ factor). This assumes that you are using a LIC (locally inertial coordinates system). You are gently guided to this formula in MTW Gravitation Ex. 2.5 page 65 and it coincides with your eq. 1.
So  eq. 1 is simply the gamma corrected projection of the equation $$ v^\mu = T^\beta \nabla_\beta X^\mu$$
You can better see their relation, by observing that - since $X$ and $T$ vectors commute (they form a 2D coordinate system) - you have that
$$ v^\mu = X^\beta \nabla_\beta T^\mu$$ which represent the difference in 4-velocity on neighboring geodesics, similar to eq. 1.
You can read about this in MTW Box 11.4, for example.
