The line element $\Delta s^2$ is suppose to be an invariant of Euclidean space. In standard coordinates $\Delta s^2=\Delta x^2+\Delta y^2$ while in polar coordinates $\Delta s^2=\Delta r^2+r^2\Delta \theta^2$. Equivalently, in standard coordinates the metric is $$\begin{bmatrix}1&0\\0&1\end{bmatrix}$$ while in polar coordinates it is $$\begin{bmatrix}1&0\\0&r^2\end{bmatrix}.$$
I don't understand how to compute the length of a point. Suppose the point is $(a,b)$ in standard coordinates. It's squared length in standard coordinates is given by $$(a,b)\begin{bmatrix}1&0\\0&1\end{bmatrix}(a,b)^T=a^2+b^2$$But how do you compute the length in polar coordinates? Wouldn't it be $$(\sqrt{a^2+b^2},\arctan(b/a))\begin{bmatrix}1&0\\0&a^2+b^2\end{bmatrix}(\sqrt{a^2+b^2},\arctan(b/a))^T$$which is definitely not the same as $a^2+b^2$.
How do you get the same line element in polar coordinates versus standard coordinates?
$\newcommand{\dd}{\partial}$To say the Euclidean metric in polar coordinates is $$ \left[\begin{array}{cc} 1 & 0 \\ 0 & r^{2} \end{array}\right] $$ really means that $ds^{2} = dr^{2} + (r\, d\theta)^{2}$. In other words, if $$ v = a\, \dd_{r} + b\, \dd_{\theta} \tag{1} $$ is a tangent vector at some point $(r, \theta)$ in polar coordinates, then the Euclidean length of $v$ is $$ \sqrt{a^{2} + (rb)^{2}}. \tag{2} $$
To calculate the arc length of a curve, use (2) to calculate the curve's speed, then integrate.
For example, to calculate the length of the parametrized line segment $$ \gamma(t) = (r_{0}t, \theta_{0}),\quad 0 \leq t \leq 1 $$ joining the origin to the point with polar coordinates $(r_{0}, \theta_{0})$, note that
The velocity of gamma is $v = r_{0}\, \dd_{r}$.
The speed of $\gamma$ is $\sqrt{r_{0}^{2}} = |r_{0}|$ by (2).
The length of $\gamma$ is the integral of the speed, $$ \int_{0}^{1} \|\gamma'(t)\|\, dt = \int_{0}^{1} |r_{0}|\, dt = |r_{0}|, $$ the Euclidean distance.