C++ move semantics on returned value

In term of move semantics on return value, I heard that the best practice is to return the object as is, and don't call std::move to it, as it prevent return value optimizations. However, consider the following code.

class MyTestClass {
public:
    // intentionally implicit constructor
    MyTestClass(const std::string& s) {
        std::cerr << "construct " << s << std::endl;
    }

    MyTestClass(std::string&& s) {
        std::cerr << "move construct " << s << std::endl;
    }

    MyTestClass(int s) {
        std::cerr << "construct int " << s << std::endl;
    }
};

void testRVO() {
    auto lambda = [](int t) -> MyTestClass {
        if (t == 0) {
            std::string s("hello");
            return s;  // use move, RVO
        } else if (t == 1) {
            std::optional<std::string> s("hello");
            // or unique pointer
            // std::unique_ptr<std::string> s = std::make_unique<std::string>("hello");
            return *s;  // use copy, no RVO !!!
        } else if (t == 2) {
            std::optional<std::string> s("hello");
            return std::move(*s);  // use move
        } else {
            return 1;
        }
    };

    lambda(0);
    lambda(1);
    lambda(2);

    lambda(-1);
}

MyTestClass simulates a generic wrapper/container (think of std::variant) that can be implicitly constructed from const T& and T&&.

In the first case, I returned a string as is, and move constructor is called, which is what you would expect.

In the second case, I constructed an std::optional (or std::unique_ptr) and returns its value using the dereference operator. However, in this case, copy constructor is called instead.

In the third case, similar to previous case, I explicitly called std::move in the return statement and the move constructor is called.

I am not sure why the copy constructor is called in the second case, as the dereference operator returns a lvalue reference, which is an lvalue. The string s in first case is also an lvalue. So I would expect them to be treated the same. Is the compiler just not smart enough to tell that the std::optional would be destroyed after return anyway? Or is it because I am using implicit construction?

In practice, I would obviously want to use the third case, since copying s can be much more expensive than moving it (consider s is a larger, more complex object), but it still feels wrong to write return std::move(...);

Could someone explain why this happens? When should I use std::move in return statement?