I'm trying to visualize covariant derivatives, so I imagines my vector, $\vec{A}$, is the north pointing needle of a compass and we are walking along the surface of the Earth at constant distance from the north pole. In standard polar spherical coordinates where $\theta$ is the polar angle (measured from north) and $\phi$ is the azimuthal angle, I found the nonzero Christoffel symbols: $$\Gamma_{\theta\phi}^\phi=\Gamma_{\phi\theta}^\phi=\cot\theta,\quad\Gamma_{\phi,\phi}^\theta=-\sin\theta\cos\theta$$ Our vector only has on nonzero component: $A^\theta=-A$. And our line element is $dx=r\sin\theta d\phi$.
Using: $DA^\alpha=A^\alpha_{,\beta}dx^\beta+\Gamma_{\mu,\beta}^\alpha A^\mu dx^\beta$, we can simplify this to: $$DA^\alpha=\Gamma_{\theta\phi}^\alpha A^\theta dx^\phi$$
Question 1: When using $dx^\phi$, is this just $d\phi$ or is it the line element component $r\sin\theta d\phi$?
Question 2: This gives me that $DA^\phi=-A\cot\theta d\phi$ or $DA^\phi=-Ar\cos\theta d\phi$. My guess for my first question is that we take $dx^\phi=d\phi$ since it diverges at the poles. We see that in either case $DA^\phi$ is only zero at the equator, I initially expected it to be zero everywhere since we don't see the needle of the compass move (if we walk eastwards it always points left). Is this because we are only travelling along a geodesic (i.e., a great circle) along the equator, so the parallel transport to compare changes in $\vec{A}$ "cut corners by following the shortest path" when we aren't walking along the equator? From our perspective we imagine that $\vec{A}$ is changing because it remains constant but we aren't moving straight (i.e., along a geodesic)? Like if we are a few meters from the north pole then we can see ourselves walking in a circle around it, which we can see isn't straight even over a curved surface?