One approach (broadly sketched) would be as follows.
First show that if a maximal area is covered,
every side of the square must be touched be a circle.
Prove this by contradiction:
Assume there is a side not touched by a circle, let the circle center $C_1$ of radius $r_1$
be closer to that side than the circle center $C_2$,
and construct a circle whose radius is larger than $r_1$
and that does not overlap either the circle at $C_2$ or any side of the square.
(One construction is, take the line tangent to both circles where they touch;
that line plus sides and/or partial sides of the square form a triangle, quadrilateral,
or pentagon enclosing $C_1$; inscribe a circle in that polygon.
Show in each case that the circle inscribed in the polygon has a radius
greater than $r_1$.)
That's the hard part.
Next show that if a circle touches more than two sides, it touches all four
(and you end up with the diagram you showed).
Then assume there is a maximal-area arrangement in which
neither circle touches more than two sides.
Since all four sides must touch a circle, one pair of sides must touch one circle
and the remaining pair of sides must touch the other.
Show that the sides in each pair cannot be opposite sides, so they must be adjacent sides.
Show that if a circle touches two adjacent sides, its center lies on the diagonal
between those two sides.
Moreover, the diagonal between one pair of adjacent sides is the same as the
diagonal between the other two sides.
Hence the centers of the two circles lie on the same diagonal.
You can then easily show by contradiction that the circles must touch,
so the entire configuration can be determined by the radius of one circle.
From there you can use algebra or calculus to find the radius that gives the
maximum combined area of the circles.
