You keep on modifiying the original post. Makes it hard to keep up a coherent discussion. However, your new version of f[x] is equal to g[x], if one applies Gianluca's suggestion of using PiecewiseExpand:

f[x_] := Which[
  0 <= Mod[x, 2 Pi] < Pi, Mod[x, Pi], 
  Pi <= Mod[x, 2 Pi] < 2 Pi,-Mod[x,Pi]+Pi
]
g[x_] := Pi (1 - (-1)^Quotient[x, Pi])/2 + (-1)^Quotient[x, Pi] Mod[x, Pi]

PiecewiseExpand[f[x] - g[x], 0 <= x < 4 Pi]
(* 0 *)

That should answer your question:

But how can this be proved with Wolfram tools?

You can use PiecewiseExpand:

h[x_] := Which[0 <= Mod[x, 2 Pi] < Pi, Mod[x, Pi],
   Pi <= Mod[x, 2 Pi] < 2 Pi, Mod[-x, Pi]];
g[x_] := 
  Pi (1 - (-1)^Quotient[x, Pi])/2 + (-1)^Quotient[x, Pi] Mod[x, Pi];
PiecewiseExpand[h[x] - g[x], 0 <= x < 4 Pi]

I think Bill's point is that if f[Pi] == g[Pi] is False, then f[x] and g[x} are not equivalent. Therefore one cannot to get from f[x] to g[x] or vice versa. Typically, Plot fails to show differences between function when they occur only at isolated points. It does discrete sampling and does not check all possible points (obviously, since there would be infinitely many values to check).

You're right! Thank you. But my question is still valid about the function f in the form of

f[x_] := Which[0 <= Mod[x, 2 \[Pi]] < \[Pi], Mod[x, \[Pi]], \[Pi] <= Mod[x, 2 \[Pi]] < 2 \[Pi], -Mod[x, \[Pi]] +  \Pi]]

You're right! Thank you. But my question is still valid about the function f in the form of

f[x_] := Which[0 <= Mod[x, 2 \[Pi]] < \[Pi], Mod[x, \[Pi]], \[Pi] <= Mod[x, 2 \[Pi]] < 2 \[Pi], -Mod[x, \[Pi]] +  \Pi]]