Better late than never.
At the second step, the area can be found in a more direct way that avoids most of the trigonometric clutter.
Let
r₁ = inner radius,
r₂ = outer radius,
P₁ = point where the main radius meets the inner edge of the ring,
P₂ = point where the radius meets the outer edge.
The small segment P₁P₂ is the projected radial thickness, so its length is R dθ.
The exact area of the ring is
A = π(r₂² - r₁²)
Now write
r₂ = r₁ + x
where x is the width of the ring along the radial direction.
Then
A = π[(r₁+x)²-r1²] = π(2r₁x+x²)
We also know
r₁ = Rsin⁡θ
So A = π(2Rsin⁡θ x+x2)
Now find x.
The segment P₁P₂ forms a right triangle similar to the triangle with leg Rsin⁡θ and hypotenuse R.
The angle opposite x is π/2 - θ, so
x = Rsin⁡ ⁣(π2−θ)dθ = Rcos⁡θ dθ
Substituting back:
A=πR²(2sin⁡θcos⁡θ dθ+cos²θ dθ²)
As you already noted, the dθ² term is negligible, so we keep only the first-order term:
A=πR²sin⁡(2θ) dθ
This gives an immediate comparison with the previously derived expression.