I am working in a Lorentzian manifold $(M, g)$ (but I think the problem would be quite similar in a Riemannian manifold) and I am considering a timelike geodesic whose tangent vector field is denoted by $X$.
Is it possible to have two conjugate points on such a timelike geodesics, and at the same time having that the expansion $\Theta = g^{\mu \nu}\nabla_{\nu}X_{\mu}$ of the congruence of the geodesic does not blow up in correspondence of the two conjugate points?
In particular, if my manifold is the Schwarzschild spacetime, I can consider geodesics moving on a hypersurface of fixed angle $\varphi$ describing circular motion, where antipodal points are conjugate points (see e.g. the Schwarzschild section in here). The blow-up of $\Theta$ indeed points out the existence of conjugate points. My problem is that if I consider there geodesics with fixed $\theta$ (instead of fixed $\varphi$), the expansion $\Theta$ is identically zero according to my calculations, even though I am expecting $\Theta$ to blow up due to the existence of conjugate points.
The difference is that the integral curves of the vector field $$ \begin{align*} u = \dfrac{1}{\sqrt{1-\dfrac{3M}{r}}} \left( \dfrac{\partial}{\partial t} + \sqrt{\dfrac{M}{r^3}} \dfrac{\partial}{\partial \theta}\right) \end{align*}$$ used in your reference are geodesics for all fixed $\varphi$ (and fixed $r$).
But if you consider a vector field like $$ v = \dfrac{\partial}{\partial t} + \sqrt{\dfrac{M}{r^3}} \dfrac{\partial}{\partial \varphi}, $$ this is tangential to a geodesic only for $\theta = \pi/2$. Rather the integral curves of $v$ describe test particles moving along something like "lines of latitude" in Schwarzschild spacetime. These lines do not diverge or converge, hence $\Theta$ is zero.