First let's change the matrix $H$ to $H=\frac{1}{2}\left(\begin{array}{cc}I_2 & I_2 \\ I_2& P_2\end{array}\right)$, where $P_2$ is a 2x2 permutation matrix. This swapping of the rows of $H$ won't affect the singular values of the matrices. Now lets consider what affect the iteration has on a matrix of the form $\left(\begin{array}{cc}X &x \\ X &y\end{array}\right)$, where $X$ has $n$ columns. At the next iteration we will have $$ \frac{1}{2}\left(\begin{array}{cccc}X &x & X &x \\ X &y& X & y\\ X&x &X & y\\ X & y& X & x\end{array}\right)$$ To account for the linear dependence of the columns, we multiply on the right by $$\left(\begin{array}{ccc}\frac{1}{\sqrt{2}}I_n & 0 &0 \\ 0 & 1& 0 \\ \frac{1}{\sqrt{2}}I_n & 0 &0 \\ 0 & 0& 1\end{array}\right)$$ The resulting matrix is $$ \frac{1}{2}\left(\begin{array}{ccc}\sqrt{2}X &x &x \\ \sqrt{2}X &y& y\\ \sqrt{2}X&x & y\\ \sqrt{2}X & y& x\end{array}\right)$$ This accounts for why the rank of the matrix increases by one with each iteration. Also note that if $$x=y$$ then the rank will always be one. If we do this reduction at each step, you find that for the starting $K_1=[1;0]$ the matrix $G=K_L^TK_L$ appears to be in the limit a Hankel matrix plus a diagonal matrix with diagonal entries $G_{ii}=(\sqrt{2})^{-2i})$ and off diagonal entries $G_{ij}=(\sqrt{2})^{-i-j-2}$. This gives you a very nice matrix but it is only for this particluar starting vector. If you use a starting vector of $K_1=[1 ;1]/\sqrt{2}$ then $K_L$ will always have rank one and have as it's largest singular value one.
Based on the observation above, consider $$K_1=\frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1\\ 1 & -1 \end{array}\right)\left(\begin{array}{c}\alpha \\ \beta\end{array}\right)$$ such that $$K_1$$ has a 2-norm of one. Applying the iteration to this matrix we have $$K_2=\left(\begin{array}{cc}Qv &Qv\Qv &P_2 Qv \end{array}\right)