Question about "the squares"

in ANY case, if the white king is crossing the edge of the big green square, the pawn is not able to be promoted.

Is this statement correct??

I think you mean the black king...because the box is around the white pawn.

How about telling us whose move it is?

White is a rook and pawn up, so the problem should be trivial but I am assuming you want to know what happens if white just sacrifices his extra rook and pushes the b pawn.

Then, the box does matter.

The statement is mostly correct, but it's not precisely stated. If the black king is on or within the big green square, the pawn is not able to be promoted without assistance.

Of course the "rule of the square" only applies to pure king and pawn endings, but as @justbefair noted, we can get one here with 1.Rxf6+!?. It doesn't matter if it's black to move, because black can hardly evade this capture.

So: 1.Rxf6+!? Rxf6 2.Rxf6+ Kxf6 and here the white pawn cannot promote on its own. For example 3.b5?! Ke6 4.b6 Kd6 5.b7 Kc7 6.b8=Q+ Kxb8. I gave 3.b5 the dubious mark because, if the pawn can be caught it is usually better _not_ to push it, as this leads to a gain of tempo for black. Meaning, it takes black four moves to capture the b-pawn, whether or not white pushes the b-pawn. If the pawn can be caught, every push of the passed pawn is wasted. Usually! There are exceptions of geometry, though.

Nevertheless white, having three pawns versus two on the kingside, wins easily in various ways. Even after needlessly giving away the extra rook, and even after needlessly throwing away tempi with the b-pawn, white wins easily in the variation given with, for example, 7.f3. But 3.f3! (instead of 3.b5?!, or indeed instead of 4.b6, etc.) wins much more simply, because white still has the b-pawn!