There is a difference between the projective plane as an object of algebraic geometry and as a combinatorial construction (=Steiner system) that has not been adequately explained. The former makes sense over any field - the latter only over a finite field (together with the speculations that no field may be necessary for the required construction to exist).

Given any field $F$ we can construct the set $\mathbb{P}^2(F)$ in the usual way: it is the set of equivalence classes $S/\sim$ of $F^3\setminus\{(0,0,0)\}$, where two non-zero vectors $x=(x_1,x_2,x_3)$ and $y=(y_1,y_2,y_3)$ of $F^3$ are called equivalent, denoted $x\sim y$, if there is a scalar $\lambda\in F^*$ such that $y_i=\lambda x_i$ for all $i=1,2,3.$

The "points" of $\mathbb{P}^2(F)$ are equivalence classes, but to identify them we need to use coordinates. To that end we need a set of representatives - one from each equivalence class. A popular way of achieving that is to select $\lambda$ in such a way that the last non-zero coordinate is equal to $1$. After all, if $x_3\neq0$, we have (with $\lambda=1/x_3$)
$$
(x_1,x_2,x_3)\sim (\frac{x_1}{x_3},\frac{x_2}{x_3},1).
$$
We can equally well choose to scale the first non-zero coordinate to be equal to one, if so desired.

If $F$ is finite, say $|F|=q$, this means that there are $q^2+q+1$ elements in $\mathbb{P}^2(F)$:

- $q^2$ equivalence classes with representatives $(x,y,1)$, $x,y\in F$ arbitrary.
- $q$ equivalence classes with representatives $(x,1,0)$, $x\in F$ arbitrary.
- A single equivalence class with representative $(1,0,0)$.

So in a way $\mathbb{P}^2(F)$ is the union of a "usual" (affine) plane, a line and a point. If $F=\{0,1,2\}=\mathbb{Z}/3\mathbb{Z}$, then the $9+3+1=13$ elements are (the classes of) $P_1=(0,0,1)$, $P_2=(0,1,1)$, $P_3=(0,2,1)$, $P_4=(1,0,1)$,
$P_5=(1,1,1)$, $P_6=(1,2,1)$, $P_7=(2,0,1)$, $P_8=(2,1,1)$, $P_9=(2,2,1)$, $P_{10}=(0,1,0)$, $P_{11}=(1,1,0)$
$P_{12}=(2,1,0)$ and $P_{13}=(1,0,0)$. The equivalence relation means that we equate, for example, the point $(2,1,2)$ with $2(2,1,2)=(1,2,1)=P_6$ and the point $(1,2,0)$ with the poin $2(1,2,0)=(2,1,0)=P_{12}$. In the field of $q$ elements there are $q-1$ non-zero constants, so in general $q-1$ points of $F^3$ form a single equivalence class. These are the non-zero points of a 1-dimensional subspace of $F^3$, and a useful point of view is to think of points of $\mathbb{P}^2(F)$ as lines through the origin in $F^3$.

To get the combinatorial design called the projective plane we need to also specify certain subsets of $\mathbb{P}^2(F)$, called "lines". These are constructed with the following recipe. Let $U\subset F^3$ be any 2-dimensional subspace. Then the line $L_U$ consists of the equivalence classes of the non-zero points of $U$. As $U$ is a subspace, it is closed under scalar multiplication. Therefore $U$ is a union of equivalence classes. If $|F|=q$, there are $q^2-1$ non-zero vectors in $U$, and these are divided into $(q^2-1)/(q-1)=q+1$ equivalence classes. Thus the line $L_U$ will have $q+1$ "points".

Continuing the example with $F=\mathbb{Z}/3\mathbb{Z}$ we see that, if $U$ is the span of the two (linearly independent) vectors $x=(1,2,1)$ and $y=(0,1,2)$ then the non-zero points in $U$ are $x,2x=(2,1,2)$ (both in $P_6$), $y,2y=(0,2,1)$ (both in $P_3$),
$x+y=(1,0,0)$, $2x+2y=(2,0,0)$ (in $P_{13}$), $x+2y=(1,1,2)$ and $2x+y=(2,2,1)$ (in $P_9$). As promised, these fell into four equivalence classes, and thus
$$ L_U=\{P_6,P_3,P_{13},P_9\}.$$

You can do the same construction for all the $13$ 2-dimensional subspaces of $F^3$, and end up with $13$ lines, four points on each. Work out a few of those by hand to gain some fluidity.

The characteristic combinatorial properties of this construction are: 1) any pair of lines intersects at exactly one point, 2) given any two points there is a unique line containing them. These then follow from results of linear algebra: 1) any two 2-dimensional subspaces of $F^3$ intersect in a 1-dimensional subspace, 2) any two 1-dimensional subspaces of $F^3$ span a unique 2-dimensional subspace.

Hopefully this clears up some of the fog.