This is not an answer, just a counterexample. In addition to your definition of f[x1,y1], let's define h[x1,x2,y1,y2]

ClearAll[f, h]
f[x1_, y1_] := Sinh[2 x1]/(Cosh[2 x1] + Cosh[2 y1])
h[x1_, x2_, y1_, y2_] := f[x1, y1] + f[x2, y2]

How should be view g[2x,y]? Should it be the same as h[x,x,y,0]?

h[x, x, y, 0]

$$\frac{\sinh (2 x)}{\cosh (2 x)+\cosh (2 y)}+\frac{\sinh (2 x)}{\cosh (2 x)+1}$$

Or, is it h[2x,0,y,0]?

h[2 x, 0, y, 0]

$$\frac{\sinh (4 x)}{\cosh (4 x)+\cosh (2 y)}$$

It seems the $G(x,y)$ you are asking for is not well defined for this particular (nonlinear) choice of f[x1,y1].

This is not an answer, just a counterexample. In addition to your definition of f[x1,y1], let's define h[x1,x2,y1,y2]

ClearAll[f, h]
f[x1_, y1_] := Sinh[2 x1]/(Cosh[2 x1] + Cosh[2 y1])
h[x1_, x2_, y1_, y2_] := f[x1, y1] + f[x2, y2]

How should one view g[2x,y]? Should it be the same as h[x,x,y,0]?

h[x, x, y, 0]

$$\frac{\sinh (2 x)}{\cosh (2 x)+\cosh (2 y)}+\frac{\sinh (2 x)}{\cosh (2 x)+1}$$

Or, is it h[2x,0,y,0]?

h[2 x, 0, y, 0]

$$\frac{\sinh (4 x)}{\cosh (4 x)+\cosh (2 y)}$$

It seems the $G(x,y)$ you are asking for is not well defined for this particular (nonlinear) choice of f[x1,y1].