How is this format a line integral? Here is the format of some line integrals provided by a textbook I was reading:

I was looking at the last one and noticed that a lot of line integrals are put in that format. However, to me, it looks like we are simply just solving a normal integral (area under the curve) for some function that changes due to x and then we solve a completely different normal integral with a different function who changes due to y. Ideally, it looks like we're solving two unrelated integrals for two unrelated functions in terms of how one changes due to x and the other to y and the only thing related between them is that they're between the same boundaries A and B.
I was wondering how this translates to solving a line function and what each section means?
Edit: To simplify for anybody passing by. Are the f1 and f2 function simply the "height" of the contour? As in, I could define a contour from A to B and the area under that contour (i.e. the height) would be defined by f1 and f2?
 A: Motivation
In the last integral
$$
f_1 (x, y) dx + f_2 (x, y ) dy
$$
is a $1$-form. In fact, it is the most general $1$-form that can exist over $2$ variables.
A $1$-form is something that "eats" a vector and "spits out" a real number. Here's what happens when our one form eats the vector $(a,b)$:
$$
[f_1( x,y) dx + f_2 (x,y) dy](a,b) = f_1 (x,y) a + _2(x, y) b
$$
We can think about integration of $1$-forms by going back to the Riemann sum notion of an integral.
Recall that if $a = x_0 < x_1 <\cdots < x_n = b$ is a partition of $[a,b]$, then
$$
\sum_i f(x_i) \Delta x_i \approx \int_a^b f(x) dx
$$
Now we want to integrate over a curve $c : [a,b] \to \mathbb{R}^2$. Using the Riemann sum idea, we take the partition $a = t_0 < \cdots < t_n = b$ and write something like:
$$
\int_c f(x,y) ds \approx \sum_i f(c(t_i) ) [c(t_{i + 1}) - c(t_i)]
$$
But there's a problem. The integral on the left should give us a real number, but the sum on the right is going to give us a vector (if it converges at all). But wait! We have something that turns vectors into real numbers: our $1$-form.
We try instead:
\begin{align}
\int_c & f_1 (x,y) dx + f_2 (x,y) dy 
\\& \approx \sum_i [f_1 (c(t_i)) dx + f_2 (c(t_i)) dy] [c(t_{i + 1}) - c(t_i)]
\end{align}
To see how we can rewrite this in a more computationally friendly way, make the approximation:
$$
c(t_{i + 1}) - c(t_i) \approx c'(t_i) \Delta t_i = (c_1'(t_i) , c_2 ' (t_i)) \Delta t_i = (c_1 ' (t_i) \Delta t_i , c_2 ' (t_i) \Delta t_i)
$$
where we've written $c'(t_i) = (c_1 ' (t_i) , c_2 ' (t_i))$. Plugging this into our integral approximation, we get:
\begin{align}
\int_c & f_1 (x,y) dx + f_2 (x,y) dy 
\\& \approx \sum_i [f_1 (c(t_i)) dx + f_2 (c(t_i)) dy] [c(t_{i + 1}) - c(t_i)]
\\& \approx \sum_i [f_1 (c(t_i)) dx + f_2 (c(t_i)) dy] [(c_1 ' (t_i) \Delta t_i, c_2 ' (t_i) \Delta t_i) ]
\\&= \sum_i f_1 (c(t_i)) c_1 ' (t_i) \Delta t_i + f_2 (c(t_i)) c_2 ' (t_i) \Delta t_i
\\&= \sum_i f_1 (c(t_i)) c_1 ' (t_i) \Delta t_i + \sum_i f_2 (c(t_i)) c_2 ' (t_i) \Delta t_i
\\&\approx \int_a^b f_1 (c(t)) c_1'(t) dt + \int_a^b f_2 (c(t)) c_2'(t) dt
\end{align}
Answer
To answer your question:
$$
\int_c f_1 (x,y) dx + f_2 (x,y) dy = \int_a^b f_1 (c(t)) c_1'(t) dt + \int_a^b f_2 (c(t)) c_2'(t) dt
$$
where $c : [a,b] \to \mathbb{R}^2$ is a curve that starts at $A$ and ends at $B$.
Commentary
Regrettably, it is still common practice to teach that $\int fdx + gdy = \int (f,g) \cdot (dx, dy)$. I'm going to encourage you to think of an object like $fdx + gdy$ as a $1$-form with its own life, not as the "dot product" $(f,g) \cdot (dx, dy)$. A $1$-form is simply something that wants to be integrated over a curve.
As for interpretation, in the $1$-form
$$
f_1 (x,y) dx + f_2 (x,y) dy
$$
you can think that $f_1$ is how heavily we weight movement in the $x$ direction, and $f_2$ is how heavily we weight movement in the $y$ direction. So if we integrate the $1$-form $f(x,y) dx$ along a path that moves only in the $y$ direction, we expect to get $0$, which we do.
A: The last one is the vector form of a line integral. Say you have a force $\vec{F}$ and you want to find out how much work $W$ it does along the path defined by $\vec{r}(t)$. Then a small work $dW$ will be given by
\begin{align}
dW &= \vec{F} \cdot d\vec{r}' \\
\implies W &= \int_{C} \vec{F} \cdot d\vec{r}'
\end{align}
where $C$ is the curve defined by $\vec{r}(t)$. This is a line integral no doubt, since you are integrating on an arbitrary path $\vec{r}(t)$. In general, $\vec{F}(t)$ can be any vector field, not necessarily only a force, and $W$ is in general the $\textbf{circulation}$ of the vector field $\vec{F}$ on the curve $C$.
Now say you are using a Cartesian coordinate system. For your question, let's confine ourselves to two dimensions though this method is completely general and will hold in $n$ dimensions.
Then you can break your vector fields into their $x$ and $y$ components.
\begin{align}
\vec{F}(t) &= f_1(x(t), y(t)) \vec{i} + f_2( x(t), y(t) ) \vec{j} \\
\vec{r}(t) &= x(t) \, \vec{i} + y(t) \, \vec{j} \\
\implies \vec{r}'(t) &= x'(t) \, \vec{i} + y'(t) \, \vec{j} \\
\end{align}
Therefore the dot product $\vec{F}(t) \cdot \vec{r}(t)$ will be,
\begin{equation}
\vec{F}(t) \cdot \vec{r}(t) = f_1(x(t), y(t)) \, x'(t) + f_2(x(t), y(t)) \, y'(t)
\end{equation}
Finally, the circulation will be
\begin{align}
W &= \int_{C} \vec{F} \cdot d\vec{r}' \\
  &= \int_{t_A}^{t_B} \vec{F}(t) \cdot \vec{r}'(t) dt \\
  &= \int_{t_A}^{t_B} f_1(x(t), y(t)) \, x'(t) dt + f_2(x(t), y(t)) \, y'(t) dt \\
  &= \int_{A}^{B} f_1(x, y) \, dx + f_2(x, y) \, dy \\
\end{align}
where in the last step I've used the substitution $x'(t) dt = dx$ and $y'(t) = dy$ to remove the parameter $t$. If you look at the equation it is precisely what is mentioned as your last example.
This is the form of a line integral expressed in the form of ordinary integrals. Both the integrals separately are just regular integrals, but their combination will represent a line integral. Can you think how to do this for $n$ dimensions?
