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You simply parametrize the line or curve using one variable, and the transformation using another.

Let's say you have a curve $\vec{s}(u)$ for $0 \le u \le 1$, say $$\vec{s}(u) = (1-u) \vec{s}_0 + u \vec{s}_1 = \vec{s}_0 + u ( \vec{s}_1 - \vec{s}_0 ) = \left [ \begin{matrix} x_0 \\ y_0 \end{matrix} \right ] + u \left [ \begin{matrix} x_1 - x_0 \\ y_1 - y_0 \end{matrix} \right ]$$ ie. a straight line segment from $\vec{s}(0) = \vec{s}_0$ to $\vec{s}(1) = \vec{s}_1$.

Let's say the transformation is parametrized using $0 \le v \le 1$, and describes counterclockwise rotation around origin by $v\cdot 180°$: $$\mathbf{R}(v) = \left [ \begin{matrix} \cos(\pi v) & -\sin(\pi v) \\ \sin(\pi v) & \cos(\pi v) \\ \end{matrix} \right ]$$

The transformed curve, say $\vec{t}(u, v)$, is the transformation applied to the curve, i.e. $$\vec{t}(u, v) = \mathbf{R}(v) \vec{s}(u)$$ That's it. It really is that simple.

You can obviously expand $\vec{t}(u, v)$: $$\begin{aligned} \vec{t}(u, v) &= \mathbf{R}(v) \vec{s}_0 + u \mathbf{R}(v) \left( \vec{s}_1 - \vec{s}_0 \right) \\ \; &= \left [ \begin{matrix} x_0 \cos(\pi v) - y_0 \sin(\pi v) \\ x_0 \sin(\pi v) + y_0 \cos(\pi v) \\ \end{matrix} \right ] + u \left [ \begin{matrix} (x_1 - x_0) \cos(\pi v) - (y_1 - y_0) \sin(\pi v) \\ (x_1 - x_0) \sin(\pi v) + (y_1 - y_0) \cos(\pi v) \end{matrix} \right ] \\ \; &= \left [ \begin{matrix} x_0 \cos(\pi v) - y_0 \sin(\pi v) + u (x_1 - x_0)\cos(\pi v) - u (y_1 - y_0) \sin(\pi v) \\ x_0 \sin(\pi v) + y_0 \cos(\pi v) + u (x_1 - x_0) \sin(\pi v) + u (y_1 - y_0) \cos(\pi v) \\ \end{matrix} \right ] \\ \; &= \left [ \begin{matrix} \left(x_0 + u ( x_1 - x_0 ) \right )\cos(\pi v) - \left(y_0 + u (y_1 - y_0)\right)\sin(\pi v) \\ \left(x_0 + u ( x_1 - x_0 ) \right )\sin(\pi v) + \left(y_0 + u (y_1 - y_0)\right)\cos(\pi v) \\ \end{matrix} \right ] \\ \end{aligned}$$ and although it might look a bit odd, it still describes the surface. If $\vec{s}_0$ or $\vec{s}_1$ is at origin, the surface is a disk; otherwise it is a circular annulus.

There are many ways of parametrizing the same surface. It just depends on how you parametrize the curve and the transformation. For example, if $\vec{s}_0$ is at origin, we get $$\vec{t}(u, v) = \left [ \begin{matrix} u x_1 \cos(\pi v) - u y_1 \sin(\pi v) \\ u x_1 \sin(\pi v) + u y_1 \cos(\pi v) \\ \end{matrix} \right ]$$ If we additionally use $$r = \frac{u}{L}, \quad L = \sqrt{x_1^2 + y_1^2}, \quad \theta = \pi v + \operatorname{atan2}(y_1, x_1)$$ where $\operatorname{atan2}(y, x)$ is similar to $\arctan(y/x)$ but takes the quadrant of $(x, y)$ into account, we get the familiar conversion from polar to Cartesian coordinates, $$\vec{t}(r, \theta) = \left [ \begin{matrix} r \cos(\theta) \\ r \sin(\theta) \\ \end{matrix} \right ] = r \left [ \begin{matrix} \cos(\theta) \\ \sin(\theta) \\ \end{matrix} \right ]$$