Parallel transport along geodesics on a sphere The general solution when parallel transporting a vector $V$ along a curve of constant $\theta_0$ (constant latitude) around a 2-sphere is (see e.g. page 4 of these notes)
\begin{align}
V^\theta(\phi) & = A\cos(\phi\cos\theta_0)+B\sin(\phi\cos\theta_0), \\
V^\phi (\phi) & = C\cos(\phi\cos\theta_0) + D\sin(\phi\cos\theta_0).
\end{align}
When $\theta_0 = \pi/2$ (along the equator), this seems to suggest that $V^\theta$ and $V^\phi$ are always constant. Does this seem correct? 
What about when transporting through a line of constant longitude to the North pole? I found in this case 
\begin{align}
\frac{dV^\theta}{d\theta} & = -\Gamma_{\phi\phi}^\theta \frac{d\phi}{d\theta} V^\phi = 0, \\
\frac{dV^\phi}{d\theta} & = -\Gamma_{\theta\phi}^\phi \frac{d\theta}{d\theta} V^\phi - \Gamma_{\phi\theta}^\phi\frac{d\phi}{d\theta}V^\theta = -\frac{\cos\theta}{\sin\theta} V^\phi, \\
\Rightarrow \ \int^{\theta=0} \frac{dV^\phi}{V^\phi} & = -\int^{0} \frac{\cos\theta}{\sin\theta} d\theta,
\end{align}
the right hand side gives $\log\sin\theta$ and is not defined when $\theta = 0$. How is such a case normally dealt with? 
Since a line of constant longitude is also a geodesic I thought about redefining the coordinate/rotate the sphere so that this path becomes the equator, but the first two equations seem to suggest that the vector is always constant along the equator. 
I think some of my lines of reasoning are wrong, but can't quite figure out which ones. Any help is appreciated. 
Edit: Thank Ted below for the answer. It is still a bit unclear to me, I guess what confuses me is that if you look at the picture below and consider parallel transport around a loop, then we may imagine $AB$ as the equator, $BN$ as the equator in a new coordinate system, $NA$ as the equator in yet another system etc. The two equations at the top says that the components of $V$ never change, but obviously after the cycle the vector $V$ is different compared to the original vector. 
I suppose by 'the components of $V$ does not change' we really mean that the orientation of $V$ is constant relative to the geodesic, but it does not mean its orientation is the same relative to its original position. But then I'm confused as to how to represent all this mathematically. 

 A: You're correct in both cases. The latitude $\theta=\pi/2$ is a geodesic and since the coordinate frame $\partial/\partial\theta,\partial/\partial\phi$ is parallel along that curve, the coefficients of $V$ will be constant.
In the second case, the spherical coordinate system fails at $\theta=0$ and $\theta=\pi$. Note that if you take the vector $Y$ to be $\partial/\partial\theta$ initially, then $Y^\phi$ starts out, and stays, $0$ and $Y^\theta$ is identically $1$. Indeed, $\partial/\partial\theta$ is parallel along this geodesic. Your phenomenon with $Y^\phi$ occurs because the vector field $\partial/\partial\phi$ has length $1/\sin\theta$; note that the quantity $Y^\phi\sin\theta$ (which is the actual length of the component of $Y$ orthogonal to the curve) does stay constant, as you'd expect.
EDIT: In the case of the geodesic triangle you've drawn, parallel translating the vector along each of the edges, it stays constant. But the net change when you come back to the original vertex is due to the angles at the vertices of the triangle. Again, part of your problem is that the spherical coordinates coordinate system is not defined at the north pole (namely, the vertex at which you make the final comparison).
