Let K be a field with $char(K) \neq 2$. Every polynomial $f \in K[X,Y] $ has a unique representation $$ a_{11}x^2+a_{22}y^2+2a_{12}xy+ 2a_{13}x + 2a_{23}y + a_{33}$$
that can be identified by a matrix $$ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{12} & a_{22} & a_{23} \\ a_{13} & a_{23} & a_{33} \\ \end{pmatrix} $$ A is symetric, so there is $ T \in GL_3(K) $ so that $T^TAT = diag(c_1,c_2,c_3)$. The related quadratic form $ F=(x,y,z)A(x,y,z)^T $ then equals $$ c_1x^2+c_2y^2+c_3z^2 $$ Since K is algebraically closed, we can suppose that $c_i = 0$ or $c_i = 1$.
How does that work? I don't see how a zero in every polynomial leeds to this conclusion.
To be fair, the author states that we can only relate this to our polynomial f if we chose z = 1, but as far as I understand the argument above still holds.
As commented by Jan-Magnus Økland: X,Y and Z can be substituted by $ x'= \sqrt{c_1}X $ and so on. We get $K[X,Y,Z] \cong K[x',y',z']$. Since K is algebraically closed, we can do this with arbitrary constants $c_i \in K$.