How can Mathematica be made to express Arg[x+Iy] in terms of ArcTan[f(x,y)] with a function f ?

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How did you type the squares? The full form of your x^2 gives x\262. The two expressions are not totally equivalent on the negative real line:

{Arg[x + I  y], 2 ArcTan[y/(x + Sqrt[x^2 + y^2])]} /. {x -> -1, 
  y -> 0}

I suppose the two functions Arg and ArcTan are meant for complex and real usage, not a mixture. However, it is disappointing that the simplification does not work.

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  1. I changed the squares into xx and yy and got no change.
  2. Is there really something like a mixed complex and real usage of functions Arg and ArcTan ? My view on the matter is as follows: In my ‘non-evaluable’ example each function gets an argument of a type which is allowed for it and returns a value which is determined solely by the function’s name and the value of the argument. No function call elsewhere can have an influence on the return value. How the function calls work, does not depend on for what kind of usage they are meant - it is defined by the syntax of the language. As I wrote: my view. Can you agree?
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Yes, it is disappointing. This gets rid of Arg in terms of ArcTan:

ComplexExpand[Arg[x + I  y] - 2  ArcTan[y/(x + Sqrt[x^2 + y^2])], 
 TargetFunctions -> {Re, Im}]
FullSimplify[%, x \[Element] Reals && y \[Element] Reals && y != 0]