I convince myself here of the form of a projective transformation of the plane expressed in Cartesian coordinates.

Generators of projective transformations:
(x', y') = A(x, y)
where A is a real 2 by 2 matrix.
x' = x + a
The two above generate all of the affine transformations.
Perhaps those two can best be combined as
(x, y) → (ax + by + c, dx + ey + f)
This correctly counts the 6 degrees of affine freedom. To be reversible ae ≠ bd.

Lets see what happens it we introduce one projective transformation which is not affine. Consider two planes in 3 space:

The triples above are conventional 3D coordinates for 3 space.

Consider the projection thru the origin (0, 0, 0). Consider a different point P in 3 space with integer coordinates (X, Y, Z). The line thru the origin and P pierces the plane 1 at (1, Y/X, Z/X). That line pierces the plane 2 at (X/Y, 1, Z/Y). (1, Y/X, Z/X) → (X/Y, 1, Z/Y).

In local coordinates (Y/X, Z/X) → (X/Y, Z/Y).
Equivalently (x, y) → (1/x, y/x) is the same transformation.

Stirring with (x, y) → (x+a, y) gives
(x, y) → (1/(x+a), y/(x+a)) In the other order we get
(x, y) → (1/x + a, y/x)
adding (x, y) → (y, x) yields
(x, y) → (1/y + a, x/y) and
(x, y) → (y/x, 1/x + a).

I think this converges at
(x, y) → ((dx + ey + f)/(ax + by + c), (gx + hy + i)/(ax + by + c)).
The line ax + by + c = 0 is the inverse image of the line at infinity. If k is a non zero real then multiplying a, b, c,... i each by k produces the same transformation. This indicates 8 degrees of freedom in agreement with the projective square. These transformations map rational points onto rational points when the coefficients are rational.

In plane 1(x, y) = (1, x, y) whereas in plane 2(x, y) = (x, 1, y) We have the transformation: We denote by 1(x, y) the point (1, x, y) which provides conventional coordinates in plane 1. Likewise 2(x, y) = (x, 1, y) names the points in plane 2.