Well, the assignment of #x# and #y# axes is arbitrary... (are you using a right-handed or left-handed coordinate system?)
So, let's choose #m_l = 1# for #p_x# so that #m_l = -1# for #p_y#. That means #m_l > 0# for orbitals involving the #x# but not #y# axis, and #m_l < 0# for orbitals involving the #y# but not #x# axis.
We also choose #m_l = -2# for #d_(x^2-y^2)# so that #m_l = 2# for #d_(xy)#. These are related since they are both on the #xy# plane...
This corresponds to what is seen here.
#d_(z^2)# and #p_z# are always #m_l = 0# by default (they are on the principal rotation axis), so with our choices of #m_l# for #p_x#, #p_y#, #d_(x^2-y^2)#, and #d_(xy)#, we get #m_l = 1# for #d_(xz)# and #m_l = -1# for #d_(yz)#.
Hence, with our choices,
#ul(l = 1)#:
#m_l = {-1,0,+1}#
#harr {p_y, p_z, p_x}#
#ul(l = 2)#:
#m_l = {-2,-1,0,+1,+2}#
#harr {d_(x^2-y^2), d_(yz), d_(z^2), d_(xz), d_(xy)}#
