Subscript may format like Derivative. But it is an "inert" wrapper basically, conveying no mathematical content. How that plays out might be seen from looking at FullForm.

In[81]:= FullForm[fun = D[f[x, y], {y, 2}]]

(* Derivative[0, 2][f][x, y] *)

In[82]:= FullForm[
 F = Subscript[f, 1][x, y] + \[Delta]^2 Subscript[f, 2][x, y]]

(* Plus[Subscript[f, 1][x, y], 
 Times[Power[\[Delta], 2], Subscript[f, 2][x, y]]]*)

In[83]:= FullForm[fun /. f -> F]

(* Derivative[0, 2][
  Plus[Subscript[f, 1][x, y], 
   Times[Power[\[Delta], 2], Subscript[f, 2][x, y]]]][x,y] *)

This will handle explicit derivatives:

fun /. Derivative[j_, 0, 0][f][x, y, t] -> Derivative[j - 1, 0, 0][g][x, y, t]

It won't take integrals though (should that be needed).

Thanks! Very helpful. Yes, there should be a d^2. I've edited it now.

Unless I missed something there should be d^2 next to the first term, or?

Here is what you can do:

fun /. f -> Function[{x, y}, d^2 g[ d x, d y]] 

or, less verbose

fun /. f -> ( d^2 g[ d #, d #2]& )

Why did your examples fail?

Nothing prevents D[f[x,y],x] from evaluation so it evaluates to Derivative[1, 0][f][x, y].

• This is why your first example fails, it does not much it.
• The second example does not work because you are replacing f, which is an operator with a product of \[Delta]^2 g where g was supposed to be an operator but how could MMA know it.

• Same story with the third example, additionaly, if you have a list of list of replacements you will get a list. You should provide a flat list of rules to get one result only.

• If you would've applied a flat list in the forth example, you'd get a correct result. But writing rules for each derivative is not something one would like to do. My solution avoids that because if you replace f in Derivative[...][f] with a function, Derivative will take care about applying the chain rule.