Yes, there are 12 orbitals, but they are typically drawn differently from what you have. It is generally assumed that the oxygen $2s$ orbitals lie too low in energy to be involved in bonding, so you have an $a_1$ and $b_1$ orbital at low energy comprised of those two $2s$ orbitals in the same phase ($a_1$) or opposite phase ($b_1$) with no $\ce{Cl}$ orbital involved. You have only included the $b_1$ of this pair. The $a_1$ replaces the $a_1$ orbital you have drawn with oxygen $p_y$ orbitals. Then your $a_1$ pair comprised of $s$ orbitals from all three atoms would instead be comprised of chlorine $3s$ and oxygen $p_y$.
You can read a description of this in Walsh's original paper[1] on AB2 non-hydride molecules.
He also assumes that in bent molecules like $\ce{ClO2}$ there is significant mixing of the $s$ and $p_z$ orbitals on the central atom so that in the $\sigma/\sigma^*$ and $\pi/\pi^*$ $a_1$ groups, the orbital on $\ce{Cl}$ is similar to an $sp$ hybrid in both cases rather than an $s$ orbital in one pair and $p$ in the other.
The relative energy levels of the orbitals for $\ce{ClO2}$ have been reported by Wang and Wang[2].
[1] Walsh, A.D. J. Chem. Soc. (1953) 2266-88.
[2] Wang, X.-B. and Wang, L.-S. J. Chem. Phys. (2000) 113:10928-33.
