How does std::move harms the original value?

In the following examples from cpp reference:

#include <iostream>
#include <utility>
#include <vector>
#include <string>

int main()
{
    std::string str = "Hello";
    std::vector<std::string> v;

    // uses the push_back(const T&) overload, which means 
    // we'll incur the cost of copying str
    v.push_back(str);
    std::cout << "After copy, str is \"" << str << "\"\n";

    // uses the rvalue reference push_back(T&&) overload, 
    // which means no strings will be copied; instead, the contents
    // of str will be moved into the vector.  This is less
    // expensive, but also means str might now be empty.
    v.push_back(std::move(str));
    std::cout << "After move, str is \"" << str << "\"\n";

    std::cout << "The contents of the vector are \"" << v[0]
                                         << "\", \"" << v[1] << "\"\n";
}

Using std::move may cause the original value be loosed. To me, it looks like

v.push_back(std::move(str))

causes a new member v[1] being created. Then,

&v[1]=&str

But why should it damage the value in str? It does not make sense.

There are many complicated tutorials about std::move which are harder than my own question to understand.

Could any one please write

v.push_back(std::move(str))

in its equivalent using c++03?

I look for an explanation whose understanding is easy and do not contain prerequisites such as x-value , static_cast, remove_reference, etc, as they themselves require to understand std::move first. Please avoid this circular dependency.

Also these links do not answer my question: 7510182, 3413470

Because I am interested in knowing how str is harm and not what happens to v[1].

Psudo code is also welcome as far as it is as simple as c++03.