How to check what type is currently used in union?

Depending on the application, if it is a short lived object you may be able to encode the type in the control flow, ie. have separate blocks/functions for both cases

  struct value {
      const char *name;
      myunion u;
  };

  void throwBall(Ball* ball)
  {
     ...
     struct value v;
     v.name = "Ball"; v.u.b = 1.2;
     process_value_double(&v);      //double
     struct value v2;
     v2.name = "Age";
     v2.u.a = 19;
     check_if_can_drive(&v2);       //int
     ...
  }

  void countOranges()
  {
       struct value v;
       v.name = "counter";
       v.u.a = ORANGE;
       count_objects(&v);          //int
  }

Is there any better solution?

No, the solution that you showed is the best (and the only) one. unions are pretty simplistic - they do not "track" what you've assigned to what. All they do is let you reuse the same memory range for all their members. They do not provide anything else beyond that, so enclosing them in a struct and using a "type" field for tracking is precisely the correct thing to do.

There is no way to directly query the type currently stored in a union.

The only ways to know the type stored in a union are to have an explicit flag (as in your mystruct example), or to ensure that control only flows to certain parts of the code when the union has a known active element.

C does not automatically keep track of which field in a union is currently in use. (In fact, I believe reading from the "wrong" field results in implementation defined behavior.) As such, it is up to your code to keep track of which one is currently used / filled out.

Your approach to keeping a separate 'uniontype' variable is a very common approach to this, and should work well.