I am more of a physicist than a mathematician, but this question is properly mathematical rather than physical, even though it is motivated by a physical application; please assume mathematical ignorance (and apologies for any abuse of notation and terminology).
Take a standard general relativist's spacetime manifold $M$, i.e. a semi-Riemannian/Lorentzian paracompact Hausdorff manifold such that $\forall p\in M$ there is an unique (timelike) geodesic $\gamma_{p,\vec{v}}$ through $p$ with a given velocity $\vec{v}$ (i.e. for each $\vec{v}\in T_{p}M$).
The specific question is this.
If, after unspecified cut & join operations on $M$, one finds that for some $p\in M, \vec{v}$ there are at least two (timelike) originally geodesic curves through $p$:
What properties of the manifold required for uniqueness theorems to apply (e.g Picard-Lindelöf/Cauchy-Lipschitz) are no longer satisfied (and specifically, why), and thus where do the uniqueness arguments break down?
Are there some properties which necessarily have been violated, or are there one or more merely sufficient violations?
I am most interested in the answers for excisions limited to the interiors of 4D regions within $M$, but pedagogical examples of other cases are also welcome.
1 Aug 16 - Additional Illustrations & Commentary
For clarification, as requested.
For definiteness: initial manifold is Minkowksi space and $\gamma_{L}, \gamma_{R}$ are geodesics with $\vec{v}=0$ in the chosen frame, but the question remains for Lorentzian manifolds in general, as originally described; two spatial dimensions suppressed.
Specify two tubes as sphere $\times$ the solid lines; excise the interiors (stippled regions) and identify the surfaces of the spheres as indicated at equal $t$; after the cut and glue*...
Fig 1: The walls of the excised regions are not parallel; geodesics $\gamma_{L}, \gamma_{R}$ are initally distinct, but the identification of $p, q$ creates a Y and the geodesics can only continue on the LHS; there appear to be two geodesics through $p, q$
Fig 2: The walls of the excised regions are parallel; $\gamma_{L}, \gamma_{R}$ cross - or continue on the same side without any rationale for going one way or the other.
Thoughts on what is wrong with the resulting manifold (Previously omitted to keep the question short and clean.) In excising the interiors of the 4D regions specified, a 3D boundary is created (time x 2D spatial). The indentification stated (which is traditional in physics literature) is purely spatial therefore the boundary does not seem to be completely removed.
Whether, if true, this affects the differential structure and in such a way as to nullify the geodesic nature of one or more curves I don't know: in Minkowski space the metric is constant and it is hard for me to see how the cut & glue is problematic from this perspective, or in the more general case if derivatives are defined for closed intervals on $\mathbb{R}$, which they can be, though I can see there might be a problem with the total derivative on $\mathbb{R}^{4}$.
In Fig 1 I do not think the solution is that the RHS line can be rotated to match the LHS because the implied homeomorphism does not, AFAICT, result in the neighbouhood $p,q$ becoming homemorphic to $\mathbb{R}^{4}$ because an extra gluing step is required. Is there a legitimate way to do this? That could be part of the answer; I don't know - hence the original question.
Fig 2 is most illustrative because at $p,q$ the tangents are parallel, so upon identification there is no need for any "rotation", but is the neighbouhood of $p,q$ homeomorphic to $\mathbb{R}^{4}$?
Conjecture
I conjecture - and am trying to prove - that the excised regions have a unique geometry if local geodesic existence and uniqueness is to be preserved through the cut and glue operation, and that geometry is some (smooth, closed, etc.) surface $\times$ a geodesic congruence; my problem with proving it is that I think I can see something's wrong if the geometry is incorrect, but I can't identify the right mathematical concepts to pin down the issue.
* I deliberately used "join" previously because I was aware of standard "cut and glue" methods and did not wish to imply any such that might have been incompatible with the motivation of the question.
Qualitatively, gluing in Riemannian or Lorentzian geometry is a rigid affair. Think of paper, or the plastic used to manufacture beverage bottles: It's flexible, but not stretchable.
A (closed) disk of paper admits a smooth flat Riemannian structure (with boundary); the exterior, i.e., the paper outside that (open) disk, also admits a smooth Riemannian structure. Their union, glued along their common boundary, admits a smooth Riemannian structure: the structure of the original flat sheet of paper before the disk was excised.
The union of two flat disks (closed, of equal radius) attached along their boundary does not acquire a smooth Riemannian structure from the geometries of the individual disks: Along the boundary, the curvature is infinite, albeit in such a way that the integral of the curvature over the boundary is finite.
Similarly, the union $M$ of two exteriors of disks (open, of equal radius) fails to acquire a smooth Riemannian structure along their common circle boundary $C$. The resulting singular space exhibits non-unique geodesics through each point of $C$: A line tangent to $C$ in one "sheet" is geodesic in any reasonable sense; each point $p$ of $C$ lies on two such lines in $M$, giving four geodesics (the lines, and the "crossovers"). Arguably, $C$ itself is geodesic, as well (e.g., it is the fixed point set of an isometry), though $C$ does consist entirely of points where the metric structure of $M$ is not Riemannian.
A square of paper admits a smooth flat Riemannian structure (with boundary). By contrast with the disk, the union of two squares attached along their boundary does acquire a smooth Riemannian structure along the edges of the squares (a $2 \times 1$ rectangle of flat paper may be folded in half!), but topological sphere obtained by attaching two squares along their boundaries still has infinite curvature at the corners, in such a way that the integral of the curvature over each vertex ($2\pi$ minus the angle incident at that vertex, a.k.a., the angular defect) is finite.
As mentioned in the comments, a sector of a disk (think Pac Man, not pie slice) can be glued into a cone. The resulting space is smooth (and flat!) except at the vertex. Lines that were parallel in the flattened sector but that meet along the "joint" always fail to be tangent.
Though each of the preceding examples gives a topological manifold (which admits a smooth structure), each paper model has "singularities", points $p$ at which the "pieces of geometry near $p$" do not comprise a smooth Riemannian disk.
The situation with Lorentzian metrics is similar, though more delicate: A Riemannian metric is isotropic to first order at each point, while a Lorentzian metric has a future-pointing light cone. That said, I believe the qualitative features of the preceding examples carry over to the Lorentzian case.
With the clarification of your diagrams (thank you!), my cone example still appears partly analogous to the left-hand picture: In the identified space, the green curve is not smooth, but instead changes direction abruptly at $p = q$. (After gluing, the two cylinder boundaries are identified; the green curve is tangent "in the future" thanks to $p$, but not tangent "in the past" thanks to $q$.)
Additionally, the left-hand identification is not spatially smooth along the cylinder boundary, for the same reason that gluing the exteriors of two open disks (of equal radius) does not yield a smooth Riemannian structure along the boundary circle.
The right-hand diagram appears to be analogous to the third example above (the glued exteriors of disks). The analogy isn't perfect, since your geodesics are timelike rather than spacelike, but (modulo my Lorentzian non-expertise) the metric in the spatial directions is not smooth along the curving cylinder.
In summary, neither proposed gluing appears to acquire a smooth Lorentzian metric, and in the left-hand diagram, the green curve is not properly tangent (in the past) to the red curve.