Assuming that the three objects you speak of are point particles initially positioned pretty much right next to each other, then all three objects will hit the earth at the same time.
From newton's law of gravitation, we have:
$F = \frac{Gm_1m_2}{r^2}$ and where $m_1$ is the mass of the Earth, and $m_2$ is the mass of an object away from the Earth (e.g. feather, or O1).
To obtain acceleration of the object (e.g. feather), we divide by mass, so:
$acceleration = \frac{Gm_1}{r^2}$
As we can see, the acceleration of the feather only depends upon the mass of the Earth. Therefore, the feather and object O1 and the elephant will all accelerate towards Earth at the same rate.
However,
O1 has a very large mass, and this will cause the Earth to accelerate towards O1. But since all three objects were initially positioned in the same spot, and all three objects are accelerating towards Earth at the same rate, the three objects remain next to each other, and therefore when the Earth reaches O1, it will reach the feather and the elephant at the same time.
The string is at a slight angle to horizontal $\theta$. It is not exactly horizontal. The slight angle is such that the tension in the string exactly counteracts gravity, $T\sin(\theta)=m g$. So, there is actually a force acting upwards that counteracts gravity, and it is supplied by the string.
You're right that if $\theta=0$ exactly, there would be a problem and the object would necessarily fall a bit.
Best Answer
In common parlance, saying that an object is "falling" implies that the object's velocity is in the direction of the ground. However, the phrase free fall is defined as applying to an object which has no forces acting on it except for gravity, a definition that has nothing to do with velocity.
If you throw a ball 3 meters above your head, your hand stops applying a force to the ball at the moment the ball leaves your hand. At that moment, there are no forces acting on the ball any more except for gravity, so it meets the definition of being in "free fall", even though at that point the direction of the ball's velocity is away from the Earth, and hence the ball isn't "falling" as that word is used in common parlance.
To answer your follow-up question, no, the ball doesn't "carry force with it" after it leaves your hand. After the ball leaves your hand, the ball has kinetic energy and momentum and an upward velocity due to the force that had been on the ball, but there is no force acting on the ball any more.