No, Marcel explained above how to calculate the interval, and it doesn't involve transporting a point to another. Parallel transport is for vectors, you don't a parallel transport a point on a manifold to another point. My guess is you are incorrectly assuming that because you can specify a point with a coordinate tuple that it is like a four-vector. A coordinate tuple is not a four-vector. — BuddyJohn Mar 7 at 10:12

@BuddyJohn A point in a manifold such as $dx^\mu$ is a place to attach a tangent (or cotangent) space where a tangent four-vector can live. I am not as clear about how to view the point since it does have a raised index which makes it look like it should be considered to be a rank 1 tensor. — sweetser Mar 7 at 20:37

The labels like $\square^\mu$ for components need to be interpreted in context. For example the labels for coordinate tuples, vectors, Dirac matrices, or some of the labels in the QCD Lagrangian, do not mean these are all similar types of objects or even the same number of components. — BuddyJohn 2 days ago

$dx^\mu$ is not a point on a manifold. In this notation, $x^\mu$ are the coordinates for a point on the manifold, and are not a tensor. And at a point $p$, $dx^\mu(p)$ is a vector in the tangent space at $p$. So $g_{\mu\nu}(p)\ dx^\mu(p)\ dx^\nu(p)$ is using the metric at $p$ to contract two vectors in the tangent space at $p$, and results in a scalar, the invariant square of an infinitesimal line element. This is sometimes just called the line element or an interval, which may have lead to the misunderstanding that it is a finite difference, but it only infinitesimal. — BuddyJohn 2 days ago

Thanks for the more explicit notation. I would also write out the two summations that are normally dropped due to the Einstein summation convention. One thing that bothers me a little is the statement that $x^\mu(p)$ would not not transform like a tensor. I would think the definition of $dx^\mu(p)$ might be something like: $lim_{\delta \rightarrow 0}x^\mu(p) - x^\mu(p + \delta) = dx^\mu(p)$. Of course $x^\mu(p)$ is not a differential line element, but it still would be a point at the start of the path. — sweetser yesterday

Why does that statement still bother you? Even if we restrict ourselves to inertial frames, consider the coordinate transformation: $(\bar{x}^0,\bar{x}^1,\bar{x}^2,\bar{x}^3) = (x^0 + a, x^1 + b, x^2,x^3)$, where $a$ and $b$ are some constants. Do you expect the components of a vectors to also change similarly? They do not. Maybe working out some problems from a textbook would help readjust your intuition. I find that helps when learning something new to help it sink it. — BuddyJohn 24 hours ago

I agree, $x^\mu(p)$ does not transform like $dx^\mu(p)$. The different is two fold: there is a delta and a limit process. Define a new operator, $DD$, which is a Discreet Difference between two points $p$ and $q$ without a limit process, so $x^\mu(p) - x^\mu(q)=DDx^\mu(p,q)$. I would think this discreet difference may transform like a tensor because at the very least the coordinate transformation above would not change $DDx^\mu(p,q)$. Convert the discrete difference operator $DD$ to the differential line element by letting $q=p+\delta$, then taking the limit of $\delta$ going to zero. — sweetser 10 hours ago