How can I fill the region where two polar plots intersect?

pp1 = ParametricPlot[Evaluate[2 {Cos[t], Sin[t]} # & /@ {1 - Cos[t], 1 + Cos[t]}], 
     {t, 0, 2 π}]

Shaded region can be obtained in two ways:

1. Post-process a one parameter ParametricPlot constrained to an appropriate region:

For example:

sh = ParametricPlot[ConditionalExpression[{2*(1 - Cos[t + π]) {Cos[t + π], 
     Sin[t + π]}, 2*(1 + Cos[t]) {Cos[t], Sin[t]}}, π/2 <= t <= 3 π/2], {t, 0, 2 π}] /. 
     Line -> ({Red, Polygon@#} &)

2. Using a two-parameter ParametricPlot constrained to the same region:

E.g.

sh = ParametricPlot[2 ConditionalExpression[v (1 - Cos[t + π]) {Cos[t + π], Sin[t + π]} + 
      (1 - v) (1 + Cos[t]) {Cos[t], Sin[t]}, π/2 <= t <= 3 π/2], 
    {t, 0, 2 π}, {v, 0, 1}, Mesh -> None, PlotStyle -> Directive[Opacity[1], Red]]

Then Show pp1 and sh together:

Show[pp1, sh, Frame -> True]

Note: In both versions, the option RegionFunction can be used instead of ConditionalExpression to constrain the plot to the desired region. That is

sh = ParametricPlot[{2*(1 - Cos[t + π]) {Cos[t + π],
       Sin[t + π]}, 2*(1 + Cos[t]) {Cos[t], Sin[t]}}, 
       {t, 0, 2 π}, 
       RegionFunction -> Function[{x, y, t, r}, π/2 <= t <= 3 π/2]] /. 
      Line -> ({Red, Polygon@#} &)

and

sh = ParametricPlot[2 v (1 - Cos[t + π]) {Cos[t + π], Sin[t + π]} + 
        2 (1 - v) (1 + Cos[t]) {Cos[t], Sin[t]}, 
        {t, 0, 2 π}, {v, 0, 1},
        RegionFunction -> Function[{x, y, t, r}, π/2 <= t <= 3 π/2],
        Mesh -> None, PlotStyle -> Directive[Opacity[1], Red]]

give the same shaded region.

It can be done in terms of the PolarPlot too. However, the kglr's shading mechanism is still needed:

p1 = PolarPlot[{2*(1 + Cos[θ]), 2*(1 - Cos[θ])}, {θ, 0, 2 π}]; 
p2 = Show[
   PolarPlot[{2*(1 - Cos[θ]), {θ, -π/2, π/2}],
   PolarPlot[{2*(1 + Cos[θ])}, {θ, π/2, 3 π/2}]}]; 
Show[p1,Graphics[{Red,Cases[p2,Line[x_]:>Polygon[x], Infinity]}],PlotRange->All]