What is the ionization steps for dibasic weak acid?

You need to consult the Ka values for both disssociation steps:
For the first dissociation:
H2S ↔ H+ + HS- : Ka = 9.1*10^-8
For the second dissociation:
HS- ↔ H+ + S 2- : Ka = 1.1*10^-12

You can use these values to calculate the % dissociation. Consider a 0.1M solution of H2S in water:

First dissociation:
Ka = [H+] [HS-] / [H2S]
We know that [H+] = [HS-] and because the dissociation of the H2S is so small we use the molarity of the original acid for [H2S]. Substitute into the equation:
9.1*10^-8 = [H+]² /0.1
[H+]² = (9.1*10^-8)*0.1
[H+]² = 9.1*10^-9
[H+] = 9.54*10^-5

% dissociation = (9.54*10^-5)/ 0.1 * 100 = 0.095% dissociated
So as you have said, the equilibrium lies far over to the left.

The second dissociation has a much smaller Ka than the first.
The [HS-] is 9.5*10^-5M and with Ka = 1.1*10^-12, the dissociation is extremely small. Once again the equilibrium is very far to the left.

Because both these dissociations are equilibrium reactions, both are reversible.