In the section 8.5.2. of the draft "An Introduction to Matrix Concentration Inequalities" written by Joel A. Tropp, the author mentioned that if $K_1$ and $K_2$ are matrices of the same dimension such that $K^{*}_1K_1 + K^{*}_2K_2 = \mathrm{Id}$, then we can choose some $L_1$ and $L_2$ making the matrix $Q$ unitary. But I am wondering why such $L_1$ and $L_2$ always exist, as we need both $K^{*}_1L_1 + K^{*}_2L_2 = \mathbf{0}$ and $L^{*}_1L_1 + L^{*}_2L_2 = \mathrm{Id}$ in order to guarantee that $Q^{*}Q = \mathrm{Id}$. So I am hoping someone can explain this point with a reasonable amount of justification.
I ended up answering my own question. The rest of the columns can be chosen to complete the basis. Using, for instance, Gram-Schmidt orthogonalization. In general, we can greedily generate vectors that are orthonormal to everything you have so far. The point here is, you already have a submatrix fixed by $K_1$ and $K_2$ — the above procedure would not work if this submatrix did not have orthonormal columns — but thankfully, by assumption, they do.