While taking Calculus III, which included some vector calculus, we defined the flux of a vector field $\mathbf{F} \colon \mathbb{R}^3 \to \mathbb{R}^3$ through a surface $S \subset \mathbb{R}^3$ as $$\iint_S \mathbf{F} \cdot \mathbf{n} \, d\sigma.$$ We were told that if we thought of $\mathbf{F}$ as the vector field defining how a fluid moves through space, the flux calculates the rate at which this fluid flows through $S$. Intuitively then, wouldn't the "units" then be something along the line of $$\frac{``\text{quantity"}}{\text{time}}.$$ So something like "The flux through $S$ is the amount of quantity that moves through it, as defined by $\mathbf{F}$, each unit of time."
However, looking at the integral, the units seem to be different. They seem to be $$\frac{``\text{quantity"}}{\text{time}} \cdot \text{area}.$$ I really don't understand what the extra area quantity is doing. If we want to know the rate at which a quantity moves through $S$, doesn't this imply that $$\frac{``\text{quantity"}}{\text{time}}$$ measures the rate at which a quantity moves through $S$ per unit of area? As if there lies in a secret dimension? Multiplying by area then gives the total flux for that small surface.
Take for example finding the flux of an electric field through a surface. The units of electric flux is $$\frac{\text{N}}{\text{C}} \cdot \text{m}^2.$$ (More surprising here is that there isn't a notion of time; but I suppose that's because previously, we thought of the vector field defining motion?) What is the extra $\text{m}^2$ factor representing? Why wouldn't the total amount of "electric field going through a surface" simply be in the units of the electric field ($\text{N}/\text{C}$) rather than including an extra "area" dimension?
The thing you integrate is actually flux density, which is a vector (or, rather, a 1-form) and has (if you ignore the fact that it is a linear form and not scalar) units $$\frac{\text{quantity}}{\text{time} \cdot \text{area}}$$ It measures the amount of quantity transported through a unit area oriented in a given direction per unit time. So when you integrate it, you sum over elementary oriented areas (act with this 1-form on the normal vector of element of the surface), and the area cancels out, leaving you with $\text{quantity}/\text{time}$.