Here’s a more or less pictorial intuition.
For $0\le\theta<2\pi$ let $R_\theta$ be the open ray $\big\{\langle r,\theta\rangle:r\ge 0\big\}$, where $r$ and $\theta$ are polar coordinates. Let $p_\theta$ be the point $\langle 1,\theta\rangle\in R_\theta\cap S^1$. Pivot $R_\theta$ about the point $p_\theta$ until it’s perpendicular to the $xy$-plane, parallel to the $z$-axis. Do this simultaneously to all of the $R_\theta$, and you end up with an open half-infinite cylinder,
$$\begin{align*}
S^1\times(-1,\to)&=\big\{\langle 1,\theta,z\rangle:z>-1\big\}&&\text{(cylindrical coords.)}\\
&=\big\{\langle x,y,z\rangle:x^2+y^2=1\text{ and }z>-1\big\}&&\text{(rectangular coords.)}\;.
\end{align*}$$
This construction leaves the $\theta$ coordinate alone and converts the $r$ coordinate into a $z$-coordinate: $z=r-1$.
Now just stretch the part of the cylinder below the $xy$-plane. There are lots of homeomorphisms that will do this; all you’re doing at this point is finding a homeomorphism between $(-1,\to)$ and $\Bbb R$.
I have no useful answer to the general question; having seen a variety of examples helps, of course. In this case the two keys for me are recognizing that removing one point and removing a closed disk leave homeomorphic spaces, and recognizing that this space that’s left is homeomorphic to an open open annulus, since it’s pretty clear that an open annulus is homeomorphic to $(0,1)\times S^1$, which in turn is homeomorphic to $\Bbb R\times S^1$.