Why is HSO3- acting as base in both the equations?

Equation 1:
H₂O + SO₂ ⇌ HSO₃⁻ + H⁺
Consider the backward reaction. HSO₃⁻ acts as a base because it accepts a proton (H⁺) to form H₂SO₃, which then decomposes to give H₂O and SO₂, i.e.
H₂O + SO₂ ⇌ H₂SO₃ ⇌ HSO₃⁻ + H⁺

Equation 2L
HSO₃⁻ + H⁺ ⇌ SO₃²⁻ + 2H⁺
Consider the forward reaction. HSO₃⁻ acts as a base because it accepts a proton (H⁺) to form H₂SO₃, which then dissociates to give SO₃²⁻ + 2H⁺.
HSO₃⁻ + H⁺ ⇌ H₂SO₃ ⇌ SO₃²⁻ + 2H⁺

Bronsted-Lowry theory ....

Equation 1:
H2O + SO2 <==> HSO3^- + H+
Equation 2:
HSO3^- + H+ <==> SO3^2- + 2H+ ............ ??? one too many H+
HSO3^- <==> SO3^2- + H+ ....................... is more like it

I would not call HSO3^- a base in either equation. In equation 1, HSO3^- is the conjugate base since it is on the product side of the equation. As written, the acid and base are on the left, and the conjugate acid and conjugate base are on the right. In the reverse reaction HSO3^- is a base.
H+ + HSO3^- <==> SO2(aq) + H2O(l) ..... there are no H2SO3 molecules in soln.

As for the second equation, HSO3^- is acting as an acid, and partially ionizes to give SO3^2- and H+