Let's compare the definitions of these three related, but distinct concepts. Let $C$ be a parametrized curve with respect to the parameter $t\in[a,b]$. Then
\begin{equation}\tag{1}
\int_C f(x,y)\,ds := \int_a^b f(x(t),y(t))\,\color{blue}{\sqrt{[x'(t)]^2+[y'(t)]^2}}\,dt
\end{equation}
whereas
\begin{align}
\int_C f(x,y)\,dx &:= \int_a^b f(x(t),y(t))\,\color{red}{x'(t)}\,dt,\tag{2}\\
\int_C f(x,y)\,dy &:= \int_a^b f(x(t),y(t))\,\color{green}{y'(t)}\,dt.\tag{3}
\end{align}
You seem to understand the geometric interpretation of (1): it is the area of the "fence" built along the curve $C$ whose height along any point $(x,y)$ on $C$ is given by $f(x,y)$. Alternatively, focus on the multiplier in blue in (1): we are weighting the integrand $f(x(t),y(t))$ by the length of the velocity vector along $C$.
On the other hand, in (2), we are weighting the integrand by only the $x$ component of the velocity vector.
In (3), we are weighting the integrand by only the $y$ component of the velocity vector.
As a simple example, consider $f(x,y)=1$.
\begin{align}
\int_C 1\,ds&=\int_a^b \sqrt{[x'(t)]^2+[y'(t)]^2}\,dt =\text{length of }C\\
\int_C 1\,dx&=\int_a^b x'(t)\,dt =x(b)-x(a)=\text{net displacement in $x$ direction as $C$ is traversed}\\
\int_C 1\,dy&=\int_a^b y'(t)\,dt =y(b)-y(a)=\text{net displacement in $y$ direction as $C$ is traversed}.
\end{align}
Draw a simple example of something like an $S$ shaped curve for $C$ and look at the three quantities above in that setting.
Edit: Here is an admittedly crude graphical interpretation of what (2) and (3) mean in the particular case when $f(x,y)=1$ (and I realize that in the picture $f(x,y)\not= 1$).
$\int_C 1\,dx$ corresponds to the dark red line on the $x$ axis while $\int_C 1\,dy$ corresponds to the dark blue line on the $y$ axis.
There are at least two worthwhile interpretations of a line integral in two-dimensions.
First, $\int_C (\vec{F} \cdot T)ds = \int_C Pdx+Qdy$ measures the work or circulation of the vector field along the oriented curve $C$. This integral is largest when the vector field aligns itself along the tangent direction of $C$. As this relates to Green's Theorem we obtain the usual form of Green's Theorem which is identified with the $z$-component of the curl a bit later in the course. For $C = \partial R$
$$ \int_{\partial R} (\vec{F} \cdot T)ds = \iint_R (\nabla \times \vec{F})_z dA $$
Second, $\int_C (\vec{F} \cdot n)ds = \int_C Pdy-Qdx$ measures the flux of the vector field emitted through the oriented curve $C$. This integral is largest when the vector field aligns itself along the normal direction of $C$. As this relates to Green's Theorem we obtain the so-called divergence-form of Green's Theorem which is related to the Divergence Theorem in due course.
$$ \int_{\partial R} (\vec{F} \cdot n)ds = \iint_R (\nabla \cdot \vec{F}) dA $$
If you want to read more, one source would be pages 357-367 of my multivariable calculus notes which were heavily influenced by Taylor's excellent calculus text.
Best Answer
The line integral $$\int_\gamma {\bf F}\cdot d{\bf r}$$ in your first displayed line has an intuitive physical meaning; the three integrals in the second displayed line don't and should not be envisaged at all.
When you have to push a cart along the curve $\gamma$ against the force field ${\bf f}$ then the total work $W$ done is roughly $$\sum_{k=1}^N {\bf F}({\bf r}_k)\cdot ({\bf r}_k-{\bf r}_{k-1})\ ,$$ where $({\bf r}_0,{\bf r}_1,\ldots,{\bf r}_N)$ is a polygonal approximation of $$\gamma:\quad t\mapsto{\bf r}(t)\qquad(a\leq t\leq b)\ .$$ It follows that $$W\doteq\sum_{k=1}^N {\bf F}\bigl({\bf r}(t_k)\bigr)\cdot \dot{\bf r}(t_k)\ (t_k-t_{k-1})\doteq \int_a^b{\bf F}\bigl({\bf r}(t)\bigr)\cdot\dot{\bf r}(t)\ dt\ .$$ Note that the integral appearing on the right hand side is a bona fide Riemann integral over the interval $[a,b]$, albeit with a complicated integrand $\Psi(t)$. Introducing the components $(F_1,F_2,F_3)$, resp. $(x, y, z)$, of the involved vectors we arrive at $$\Psi(t)=F_1\bigl(x(t),y(t),z(t)\bigr)\dot x(t)+F_2\bigl(x(t),y(t),z(t)\bigr)\dot y(t)+F_3\bigl(x(t),y(t),z(t)\bigr)\dot z(t)\ .$$ The individual summands appearing here have no meaning by themselves. Arguing purely formally one of course could write $$\int_a^b F_1\bigl(x(t),y(t),z(t)\bigr)\dot x(t)\ dt=\int_\gamma F_1(x,y,z)\ dx\ ,$$ which leads to the integrals in your second displayed line.