One place the mathematics is discussed: Points at infinity where last element in homogeneous vector is $0$?.
More geometrically, an artist's field of view may be modeled either by a sphere of rays or by a projective plane of lines through the eye. We may as well pick the unit sphere centered at the origin of Euclidean three-space.

An open hemisphere $H$ of the sphere model corresponds to a dense open set $P$ of the projective plane, and the boundary great circle $C$ of $H$ maps to the line at infinity with respect to $P$ by definition. The golden circle shown lies in the equatorial plane $Z = 0$, and the plane $P$ being visualized is $Z = -1$, which corresponds to the lower hemisphere.
Straight lines in $P$ map to the sphere by radial projection from the eye, so their images are great circles. Since great circles intersect on the sphere, the images of lines in $P$ also intersect, even if the lines are parallel as shown. By definition, parallel lines in $P$ do not intersect in $P$. The points of intersection on the sphere therefore line on the boundary great circle, a.k.a., the line at infinity with respect to $P$.
In Penrose's drawing, the artist's canvas might be the plane $Y = 1$ (not shown here). Projecting the great circles to that plane gives a pair of crossing lines, compare the railroad tracks in the linked question.
Added: The diagram below shows the same picture without the sphere, and with the projective lines shown as affine planes. To emphasize,
- Each colored affine line through the origin (Eye) represents a projective point.
- Each affine plane through the origin represents a projective line.
- We're choosing the line at infinity to be the set of gold affine lines lying in the affine plane $Z = 0$.

The plane $Z = -1$, by contrast, does not pass through the origin.
- Its points correspond to projective points: Each point $p$ of the plane $Z = -1$ determines a unique affine line $\ell$ through the origin and $p$, and $\ell$ intersects $Z = -1$ precisely at $p$ rather than being contained in the plane $Z = -1$.
- Its affine lines (the two parallel blue lines, for example, modeling the edges of Penrose's roadway) correspond to projective lines, represented by the shaded affine planes.
- The entire affine plane $Z = -1$ corresponds to the "finite" Euclidean part of the projective plane with respect to the projective line at infinity, here chosen to be the affine plane $Z = 0$.
- The two parallel affine lines in the affine plane $Z = -1$ intersect in the projective plane because the affine planes that represent them intersect along an affine line through the origin, i.e., at a point of the projective plane. The intersection of the two slanted affine planes lies in the affine plane $Z = 0$, a.k.a., the projective line at infinity. In that sense, the parallel affine/Euclidean lines intersect at infinity.