The Algorithm

The method logIn can be chunked into small steps:

public boolean logIn(String name, String password) {

   if( /* user is logged in already? */)

   for (/* every user */)
     if ( /* has name and password */)
 
   if (/* no match found */)
     for (/* every user */)
       if ( /* can find user with the given name? */ )
}

The logic is more complex than it has to be. The algorithm searches twice if a user with the name exists. A simpler algorithm could look like:

public boolean logIn(String name, String password) {
    User user = // find user by its name

    if (/* no user found */)

    if (user.isLoggedIn)

    return user.hasPassword(password)     
}

Data Structure

The method getUsers() looks like it returns a List. Since it is a list and you do not know at which index a concrete user is saved you have to search for the user.

for (User user: getUsers())
{
  if (user.toString().equals(name))

To check that a user with name does not exists you have to loop through the hole list and this could take some time if you have many users! This is also known as \$O(n)\$ which is a indicator for the time complexity a algorithm can have.

It would be much more permanentperformant if we could get a user directly without to search it which would be a time complexity of \$O(1)\$.

We can archive this goal by using a Map instead of a List:

Map<String, User> nameByUser = new HashMap<>();

nameByUser.put("TomZ", new User("TomZ", "aPassw0rt"));
// ..insert some more users

User user = nameByUser.get("TomZ");