How is the derivative of a vector field Y on a curve (almost) the covariant of Y?

This two-part problem is from O'Neill's Elementary Differential Geometry, section 2.5. It is in the second statement of the second part where I'm unclear, beginning with "Thus."

**Full problem**:

Let be a vector field defined on a region containing a regular curve . Then (i.e. composed with ) is a vector field on called the restriction of to .

1. Prove that .

2. Deduce that the straight line in Definition 5.1 (below) may be replaced by any curve with initial velocity . Thus the derivative of a vector field on a curve is (almost) .

**Definition 5.1**. Let be a vector field on , and let be a tangent vector to at the point . The the covariant derivative of with respect to is the tangent vector at the point .

The following definition is useful for part (2) since it distinguishes between a vector field and a vector field on a curve.

**Definition 2.2**. A vector field on a curve from to is a function that assigns to each number in a tangent vector to at the point .

**My attempt, and where I'm stuck**:

Part (1) was fairly straight-forward, using the definitions of covariant derivative and what it means to differentiate a composition.

Part (2) has two parts. My approach to the first part is the following, and I believe it to be correct. The idea is to define a function on curves and show that it agrees with the covariant derivative with respect to a vector at a point for all curves such that and . Part (1) can be used to show the function is well-defined and that it indeed equals the covariant.

The second part, starting at "Thus" is where I'm having trouble. It's with the use of the word "almost." To me, if is a vector field on a curve , then using the definition of covariant, straight-line or otherwise, makes no sense because is not defined (i.e., is not defined on , only ). So I thought this might be the almost part. However, what's 'almost' about it? I was thinking that perhaps given on that a vector field could be defined such that , and that then would equal the covariant of instead of . However, I don't think I can 'always' define such a .