One of the ways I like understanding things is being able to "see what's going on" so I can hypothesise intuitive results (and then rigorously prove them later).
For example, when I see $\int_C fds$, where $f:\mathbb{R^n}\rightarrow \mathbb{R}$ and $C$ is some curve in $\mathbb{R^2}$, I 'see' it as finding the area of the fence formed when I connect every point on $C$ vertically up to the surface $f$.
So using this intuition, I can hypothesise that the direction of $C$ does not matter, which is the case, so $\int_{-C} fds = \int_{C} fds$.
Likewise, when integrating a curve over a vector field, the direction does matter and you will get positive versions or negative versions of the value, depending on the direction.
My question is, could me being somewhat 'reliant' on these visualisations be 'dangerous'? For example, in elementary integration, a large majority of students (when they start off) mistake the value of the integral for always being the same as the area under the curve, when that is not necessarily the case. Is there a similar situation that could arise, except in my studies of vector calculus?
Calculus was made for physics, and you can't go wrong with thinking about integrals as a Riemann sum (an infinitely accurate sum of values of a function multiplied with certain chunks of the argument/s, whether this function returns a point on a curve or the mass at a certain place).
The FTC provides the link between antiderivatives and the area function under the function (which is the 2D riemann sum and subsequently higher dimensional Riemann sums can be broken down into the 2D sums by 'summing the sum'). Therefore any integral has an equivalent Riemann sum and hence a visual interpretation.
With path integrals, you can safely think about it as a robot travelling in that path collecting values of f(C) ds evaluated at 'every' point along the path C, and adding them together. Your visualisation is correct only if dealing with spatial coordinates, but the variables could represent anything- time, mass, density, etc.. Both can be tailored to match the Riemann sum definition which will always be valid.