Why does he call them quotients whereas in differential geometry we think of the elements as subspaces?
This becomes becomes more evident, when you want to work with the projective space as a scheme instead of a manifold or abstract variety. Let's look at the easiest case, the projective line: When $k$ is a field, there is no difference between quotients and subspaces, since everything is split. When we want to look at $\mathbb{P}^1_k(R)$ for an $R$-algebra $k$ (really you can do this for any $R$-algebra), you might be tempted to define it via subspaces, but this turns out to be insufficient, whereas quotients yield a more easy description. I recommend you look at this answer. As a caveat it should be stressed, that also you have to switch from free modules to projective modules.
How do we get the second formulation and what is $P$?
$P$ in this case is the $(r,p-r)$ parabolic. There are matrixes where the lower left $(p-r)\times r$ block is just $0$. The map goes as follows: Let's define $x\in G(p,r)$ to be the standard flag, i.e. $x=<e_1,...,e_r>$ with $e_i$ the standard basis. Given an element $g\in SL(p)$, we set $g.x:=<g(e_1),...,g(e_r)>$. You should verify that this defines a well-defined action. So you get a map
$$ SL(p)\to Gr(r,p)$$
sending $g$ to $g.x$. This is surjective, because the action is transitive. Finally you verify that $Stab_{SL(p)}(x)=P$, so you get exactly that $$ SL(p)/P\to Gr(r,p)$$ is an isomorphism. This isomorphism in fact exists in greater generality for any parabolic subgroups (don't worry if you don't know what that means). You can consult these notes for more details.
