Notation Question with Line Integrals over Vector Field

Previous Question: What is the Convention in Arc Length Parametrization?

This is a follow-up question to my previous post on line integrals going in opposite directions. At first I thought I understood it but later had another question. My text denotes the smooth parametrization of $C$ by $\mathbf r: [a, b] \rightarrow C$, and uses $u \in [0, L]$ as the arc-length parameter. The line integral of a vector field is:

$$\int_{C}\ F \cdot \mathbf T(u)\ du = \int_a^b\ F \cdot \mathbf r'(t)\ dt = \int_{C}\ F \cdot d \mathbf r$$

My best understanding of the last notation is to take the integral along $C$ as $\mathbf r$ varies. Then, the text says that if you take the line integral of the vector field in the opposite direction, you will get the negative:

$$-\int_{C}\ F \cdot d \mathbf r = \int_{-C}\ F \cdot d \mathbf r$$

I am not able to reconcile the various notations. To my previous question, Mr. John D. wrote:

$$\int_C \mathbf{F}\cdot d\mathbf{s}=\int_a^b \mathbf{F}(\mathbf{c}(t))\cdot\mathbf{c}'(t)\,dt=-\int_a^b \mathbf{F}(\mathbf{d}(t))\cdot\mathbf{d}'(t)\,dt=-\int_{-C}\mathbf{F}\cdot d\mathbf{s}$$

where $\mathbf c, \mathbf d$ are parametrizations of $C$ running in opposite directions. Thus, the equality in the middle makes sense; it follows the change of variable. What about the first and third equality? If $\mathbf s\ \text{(or}\ \mathbf r \text{)}$ parametrizes $C$ in one direction, how can you use the same thing to parametrize $C$ in the opposite direction, therefore yielding $-\int_{C}\ F \cdot d \mathbf s = \int_{-C}\ F \cdot d \mathbf s$? In my opinion, it makes better sense to write $-\int_{C}\ F \cdot d \mathbf s = \int_{-C}\ F \cdot d (\mathbf -s)$. What is your verdict on this?

My best understanding of the last notation is to take the integral along $C$ as $\mathbf r$ varies.

Not quite. Please note that $\int_C \mathbf{F}\cdot d\mathbf{s}$ is simply a notational device that by definition means $\int_a^b\mathbf{F}(\mathbf{c}(t))\cdot\mathbf{c}'(t)\,dt$ and that answers your question: $\int_C \mathbf{F}\cdot d\mathbf{s}$ means $\int_a^b\mathbf{F}(\mathbf{c}(t))\cdot\mathbf{c}'(t)\,dt$ which means at a given value of the parameter $t$, we evaluate $\mathbf{F}(\mathbf{c}(t))$ and dot that with $\mathbf{c}'(t)$, and accumulate these as $t$ varies from $a$ to $b$ (so that $\mathbf{c}(t)$ traces out $C$). In other words, at each $t\in[a,b]$, we compute the component of $\mathbf{F}(\mathbf{c}(t))$ in the direction of the parametrization velocity vector $\mathbf{c}'(t)$ and accumulate these over $a\le t\le b$ during which $\mathbf{c}(t)$ traces out $C$.

Thus, the equality in the middle makes sense; it follows the change of variable. What about the first and third equality?

The first equality is simply unpacking what the notation $\int_C \mathbf{F}\cdot d\mathbf{s}$ means. It is the definition of the notation.

To understand the third equality, read it from right to left (ignoring the outer minus signs for the time being): $$\int_{-C}\mathbf{F}\cdot d\mathbf{s}=\int_a^b\mathbf{F}(\mathbf{d}(t))\cdot\mathbf{d}'(t)\,dt$$ where the key here is that $\mathbf{d}$ is the backwards/opposite parametrization (relative to $\mathbf{c}$) of the curve $C$. This is why the integration is over $-C$ instead of $C$ here. Negate both sides and you have the third equality.

You said you understand the middle equality already.

If $\mathbf s\ \text{(or}\ \mathbf r \text{)}$ parametrizes $C$ in one direction, how can you use the same thing to parametrize $C$ in the opposite direction, therefore yielding $-\int_{C}\ F \cdot d \mathbf s= \int_{-C}\ F \cdot d \mathbf s$? In my opinion, it makes better sense to write $-\int_{C}\ F \cdot d \mathbf s = \int_{-C}\ F \cdot d(\mathbf -s)$. What is your verdict on this?

But in the notation $\int_{C}\ F \cdot d \mathbf s$, $\mathbf{s}$ is just a symbol, it is not the parametrization---look up at the beginning of my answer: $\mathbf{c}(t)$ is the parametrization!

You seem to be saying that when you reverse the direction of the parametrization (i.e. reverse the orientation on $C$), that you expect to end up negating the line integral---and that's what happens!