I finally stumbled upon the term radar to describe this use of a reference timelike worldline. A good starting resource is: Perlick, Volker. "On the radar method in general-relativistic spacetimes.", which points to a bunch of other resources.
The $r^\star$ and $\tau^\star$ are in fact the radar distance and radar time resp.
I still haven't found any name for the surface, though radar bubble suggests itself. There is a region around the reference worldline where these bubbles are globally spacelike and topologically $S^2$ (a bubble). In flat space this region is the entire space, though a gravitating object will make bubbles self-intersect at certain distance.
A radar bubble can fail to be even locally spacelike: if a null geodesic intersects the worldline twice, then the corresponding bubble would include that geodesic. A black hole can do this. It can act as a gravitational mirror, and bounce light coming in on a certain angle from a distant source back.
Addendum
Instead of the light cones themselves, the above also works using the boundary of the chronological future (past). The chronological future (past) of an event $p$ are the events that can be reached from $p$ (can reach $p$) by a timelike path. They are designated $I^\pm(p)$, and their boundaries $\delta I^\pm(p)$.
In a globally hyperbolic spacetime, these are subsets of the past or future light cone, excluding the parts where it self-intersects.
The intersection of a future and past boundary (not to be confused with the boundary of the intersection) $\delta I^+(p) \cap \delta I^-(q)$, is indeed globally spacelike, though not always topologically $S^2$.
Given a timelike path $p(\lambda)$, let's label the bubble $B_p(\lambda,\mu) = \delta I^+(p(\lambda-\mu)) \cap \delta I^-(p(\lambda+\mu))$. Then for a given $\lambda$, the union of bubbles $\bigcup_{\mu \ge 0} B_p(\lambda,\mu)$ is a globally spacelike 3-D hypersurface. This family of hypersurfaces indexed by $\lambda$ foliates the part of the spacetime reachable from the path (unlike the "instantaneous simultaneous spaces" -- even in special relativity, two instantaneous simultaneous spaces at different events of an accelerated worldline will intersect each other).
In particular this is true if the path parameter is just the proper time along the path (i.e. $\tau^\star = \lambda$ and $r^\star = \mu c$).
Spacelike, null and timelike geodesics correspond to geodesics (in a spacetime with signature $-+++$) with a tangent vector $u$ of positive, zero or negative norm, $|u| = g_{\mu\nu} u^\mu u^\nu$. It does correspond to non-accelerated motion, in fact acceleration in general relativity is deviation from geodesic behaviour :
$$a^\mu = \ddot x^\mu + {\Gamma^\mu}_{\alpha \beta} \dot x^\alpha \dot x^\beta$$
Particles moving on those curves are particles of $p^2 = -M^2 < 0$ for timelike curves (generally called massive particles or tardyons), $p^2 = - M^2 = 0$ for null curves (generally called massless particles or luxons), or $p^2 = -M^2 > 0$ for spacelike curves (generally called tachyons).
Massive particles do not necessarily end at timelike infinity, but realistically they do. The only class of particles that can reach null infinity are constantly accelerated ones, such as Rindler observers.
Best Answer
Just use the geodesic equation to evaluate the derivative along the curve of the "length" of the tangent to the curve. You'll find that it is zero, so the "length" does not change along the curve. In particular, the sign of the "length" does not change.
The geodesic equation is \begin{align} T^a \nabla_a T^b = 0, \end{align} where $T^a$ is the tangent to the curve, in whatever affine parameterization you want. The norm of the tangent (which determines whether it's timelike, spacelike, or null) is given by $T^b T^c g_{bc}$. The quantity $T^a \nabla_a$ is just the derivative along the curve, so the derivative along the curve of the norm of the tangent is \begin{align} T^a \nabla_a \left( T^b T^c g_{bc} \right) = 0. \end{align} Since the covariant derivative of the metric is zero, we can pull it out of the derivative, and expand this as \begin{align} T^a \nabla_a \left( T^b T^c g_{bc} \right) &= g_{bc} T^a \nabla_a \left( T^b T^c \right) \\ &= g_{bc} \left[ \left( T^a \nabla_a T^b \right) T^c + T^b \left( T^a \nabla_a T^c \right) \right] \\ &= 0 \end{align} To get to zero, we just apply the geodesic equation twice in that second-to-last line. Again, since this derivative along the curve is zero, the norm does not change — so the sign of the norm does not change.