Calling non-const function of another class by reference from const function

If doSomeStuff() would change the data, wouldn't that affect class B as well?

Well, not in the way a compiler checks for const correctness. A B holds a reference to an A. That object can reside anywhere, but most importantly it doesn't reside inside the B object. So modifying it does not do something with undefined behavior like changing a const object. We have a reference to a non-const object, so it's possible to modify the object via the reference. That's as far as the C++ type system cares, whether or not the object is maybe physically const.

It will probably affect the logical state of the B, but it is the responsibility of the programmer to ensure the class invariants hold. C++ will not hold your hand in that endeavor.

The original object of the class A will be changed.

When you are using a const member function then the function deals with const T *this where T is the class type.

That is data members of the object are considered constant.

For a referenced type it could look like

A & const a;

However references themselves can not be constant.

That is for example this declaration

int x;

int & const rx = x;

is invalid and does not mean the same as

const int & rx = x;

So the class B has a reference that points to a non-constant object and using the reference the object can be changed.

Compare with the following declaration of the class B

class B
{
public:
    B(A * a) : a(a) {}
    void constStuff() const { a.doSomeStuff(); }

private:
    A *a;
};

Then then a constant member function is used the data member is considered like

A * const a;

(pointers themselves may be constant) that is the pointer itself that is constant not the object pointed to by the pointer and you can not change the pointer itself but you can change the object pointed to by the pointer.