Does a frame of reference have energy in itself? When I moving object in some frame of reference, it has a positive kinetic energy. In the frame of reference where the object is stationary, it has a kinetic energy of 0. 
Is the energy difference in some way associated to the frames of reference?
 A: 
Does a frame of reference have energy in itself?

No, there's no reason to think of it that way.  When we talk about energy, it's very useful because it can be transferred between object and move from one form to another.
There's no transformation we can do that will allow us to extract energy from a frame.  So the energy differences in your scenario aren't because the energy is "in the frame".
We can imagine a frame where you are moving at $0.5c$.  Your body has a numerically large value for kinetic energy in that frame.  But you have no means to exploit that energy.
Rather than the absolute value of $KE$, you should think about how energy moves from one thing to another.  Take a pair of objects and have them collide inelastically.  Some of the $KE$ is turned into thermal energy.  In the frame where the center of mass is at rest, all the $KE$ is used in this manner.  In a frame where the center of mass is at high speed, $KE$ moves from one object to the other, but the total loss of kinetic energy into heat is identical.  In neither case do we need to consider the frame to be involved in the energy transfer.
I've added some bits from comments.

Does that mean a body that moves somehow needs a collision for the kinetic energy to manifest?

A "collision" may not be necessary, but an "interaction" with another mass is.  You can't change KE of an object without changing velocity, and you can't change velocity without changing momentum, and momentum is conserved.  If there's no interaction, then there's no change in KE.  Without being able to change it, the numerical value has no meaning.  
A large KE is only meaningful in scenarios where the KE can be reduced (giving energy to something as it slows down).  

I am thinking like this: I have a ball in my hand, without kinetic energy. I throw it, and see it moving in the air, and assume now it has kinetic energy. I do not think about a collision.

No collision, but your hand (accelerated via energy produced in your muscles) provided the energy.  If your hand was unable to interact with the ball, it would not have gained energy.

If I need to, I use the appropriate magic to create a hole through the earth, and throw the ball in, it will oscillate forever without colliding with anything. That does not mean it does not have kinetic energy, right?

It accelerates due to gravitational interaction with the earth.  You may not be able to measure the acceleration of the earth in this interaction, but it is present.  The potential energy in the gravitational system of the ball and the earth, and the kinetic energy of the earth all change as the ball falls.  After you let go of the ball, the total of all three is constant (in any inertial frame you choose).  Different frames would have different absolute values, but the energy transfer always sums up.
$$E_{tot} = KE_{ball} + KE_{earth} + GPE_{earth-ball-system}$$

That does not mean it does not have kinetic energy, right?

No, of course it has KE.  I don't think that was your question.
A: 
Is the energy difference in some way associated to the frames of
  reference?

Yes it is. The macroscopic kinetic energy of an object (the kinetic energy of an object as a whole) is its kinetic energy with respect to an external (to itself) frame of reference. 
For example, a car of mass $m$ moving with velocity $v$ relative to a person standing on the road, has a kinetic energy of $\frac{mv^2}{2}$ with respect to the person on the road.  On the other hand, its kinetic energy with respect to a person in another car driving next to it in the same direction with the same velocity with respect to the road is zero, because the velocity of the first car with respect to the second car is zero.
Hope this helps.
A: The kinetic energy is always positive.
If you are in a train and your velocity in the train is $\vec{v}_u$ and the train is moving with the velocity $\vec{v}_t$ relative to a inertial frame  then your kinetic energy is 
$$T_u=\frac{m}{2}\vec{v}^2$$
where $\vec{v}=\vec{v}_t+\vec{v}_u$
to obtain your kinetic energy you have to calculate your velocity in inertial frame.
The "moving frame" is the train with the kinetic energy 
$T_t=\frac{M}{2}\vec{v}_t^2$
and the energy difference is $T_t-T_u$.
