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I am thinking that the difference between the two covariant derivatives of a vector field along a path that starts out in direction w would be due to the different curvatures of the paths.

In my opinion, you are not quite right. The connection coefficients Gamma are a function of the point only, not of the curvature of the path. The reason for having more covariant derivatives is simply that there is some freedom in defining how "parallel transport" can be made and therefore what "covariant derivative" exactly is.

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Taking the example of the surface of a sphere, if the first path is a geodesic ...Does this make any sense?

In order to be able to speak of a geodesic, you must have defined a metric. But a metric defines uniquely the connection (i.e. the covariant derivative), see chapter 14.7 and exercise 14.28. So after having defined a metric you have lost the freedom I mentioned above and in your example there is just one covariant derivative, defined by the metrics. And the Gamma are non-zero because there is a surface curvature, independently from the path curvature.

By the way, probably we should have this discussion in the Exercise Discussion section of the forum, not in the Solutions section.