I addressed a similar question a while back; long story short, $g$ is the metric tensor field (a $(0,2)$ tensor field on the manifold $M$), and the line element $ds^2$ should be interpreted as the associated quadratic function (though sometimes people do use $ds^2$ to mean the actual metric tensor field $g$ itself, though I don't like this).
The meaning of the integral is as follows: we fix a smooth manifold $M$ and $g$ a non-degenerate symmetric $(0,2)$ tensor field on $M$ (this takes into account the Riemannian and Lorentzian case in any dimension), and suppose $\gamma:[a,b]\to M$ is a smooth curve ($C^1$ or even piece-wise $C^1$ would work), and we suppose in addition that for each $t\in [a,b]$, we have $ds^2_{\gamma(t)}[\gamma'(t)]:= g_{\gamma(t)}[\gamma'(t),\gamma'(t)]\geq 0$ (here I use $\gamma'(t)$ to denote the velocity vector, which is an element of the tangent space $T_{\gamma(t)}M$). We now consider the speed of $\gamma$, which is a function $\sigma:[a,b]\to \Bbb{R}$ defined as
\begin{align}
\sigma(t):= \sqrt{ds^2_{\gamma(t)}[\gamma'(t)]} := \sqrt{g_{\gamma(t)}[\gamma'(t),\gamma'(t)]}
\end{align}
This function is well-defined because the thing under the square root is $\geq 0$ by assumption. Also this is the square root of a continuous function of $t$, hence it is continuous, hence Riemann-integrable on $[a,b]$. So, now the meaning of the integral is
\begin{align}
\int_{\gamma}ds:= \int_a^b \sigma(t)\, dt := \int_a^b\sqrt{g_{\gamma(t)}[\gamma'(t),\gamma'(t)]}\, dt.
\end{align}
(i.e if you integrate the speed of the curve, you get its length).
If we further assume we have a coordinate chart $(U,x)$ such that the image of $\gamma$ lies entirely in $U$, then we can write this as
\begin{align}
\int_{\gamma}ds&=\int_a^b\sqrt{g_{\mu\nu}(\gamma(t))(x^{\mu}\circ \gamma)'(t)
(x^{\mu}\circ \gamma)'(t)}\, dt
\end{align}
or more concisely, $\int_{\gamma}ds= \int_a^b\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}}\, dt$, which is of course an abuse of notation since $x^{\mu}$ is being used twice with different meanings (the point is that the coordinate functions $x^{\mu}$ become functions along the curve simply by composition $x^{\mu}\circ \gamma$)... but anyway many books often write this only so its a good idea to get used to this.
For your final question, if $(U,x)$ is a coordinate chart on the manifold $M$, then saying something like $dx^{\mu}=0$ is technically an incorrect statement, since $dx^{\mu}$ is NOT the zero differential form. What is meant is that if you have a curve $\gamma:[a,b]\to M$, then the pullback form
\begin{align}
\gamma^*(dx^{\mu})= d(\gamma^*x^{\mu})= d(x^{\mu}\circ \gamma) = (x^{\mu}\circ \gamma)'\, dt
\end{align}
is the zero differential form on $[a,b]$ (and $t$ denotes the usual coordinate function on this interval).
For example, suppose the manifold is $M= \Bbb{R}^4$, and $(x^0,x^1,x^2,x^3)$ are the "cartesian coordinates". Let $\gamma:[0,1]\to M=\Bbb{R}^4$ be the curve $\gamma(t)= (\cos t e^{-t^2\int_0^t\arctan(e^{-\lambda})\,d\lambda}, 89, t^7, e^{t\ln(2+ \sin t)})$. Then, you see the $x^1$ component of $\gamma$ is constant; i.e $\gamma^1(t):= (x^1\circ \gamma)(t) = 89$ for all $t\in [0,1]$, then we often write $dx^1=0$, when we really mean $d(x^1\circ \gamma)=0$ (because $(x^1\circ\gamma)'(t)=0$ for all $t\in[0,1]$). Often times, we don't bother introducing a new letter $\gamma$ for a curve... this is just something to get used to.
