Examples from Principles of Physics (by Walker,Resnick,Halliday) will say it better:
Let us throw a tomato upward. . .as the tomato rises, the work $\mathbf{W_g}$ done on the tomato by the gravitational force is negative as the force transfers energy from the $E_K$ of the tomato. We can now finish the story by saying that this energy is transferred by gravitational force to the gravitational potential energy of the tomato-earth system. The tomato slows down, stops & then begins to fall back down. During the fall, the transfer is reversed: the work $\mathbf{W_g}$ done on tomato by the gravitational force is now positive - that force transfers energy from the gravitational potential energy of the tomato-Earth system to the $E_K$ of the tomato.
If we abruptly shove a block to send it moving rightward, the spring force acts leftward & thus does negative work on the block, transferring energy from the $E_K$ of the block to the elastic potential energy of the spring-block system. . . .
And many,many other examples. Now, why is the potential energy attributed to the system and the kinetic energy to the object itself? What is the logic/cause?
Answer
Here's an example of why this is done using a gravitational context.
Kinetic energy
The kinetic energy of a system of particles is defined to be $K_\text{tot} = \sum_i K_i$, where the sum runs over all particles in the system. This is just a definition, and choosing to use such a definition allows one to speak of the kinetic energy $K_i$ of the $i$th particle, for example, since it's just one of the terms in a sum.
Okay, that was boring.
Potential energy
Gravitational potential energy $U_\text{g} = -\frac{Gm_im_j}{d_{i,j}}$ is defined for pairs of particles.$^\text{1}$ Mathematically you can see that in the subscripts of the masses $m_i$ and the distance between the two particles $d_{i,j}$. Why? Because to determine potential energy from scratch, you must consider how two objects interact, which in classical mechanics is where forces come in. And forces themselves come in pairs.
Now, the fact that they come in pairs means one must be careful when determining the potential energy in some given situation, otherwise you might over-count. Suppose, for example, there are 4 particles with masses $m_i$, and the distance between the $i$th and $j$th objects is $d_\text{i,j}$. What, then, is the potential energy of the masses and distances are all given? The answer is $$U_\text{g}=-G\left( \frac{m_1m_2}{d_{1,2}} + \frac{m_1m_3}{d_{1,3}} + \frac{m_1m_4}{d_{1,4}} + \frac{m_2m_3}{d_{2,3}} + \frac{m_2m_4}{d_{2,4}} + \frac{m_3m_4}{d_{3,4}} \right)$$ First, note that since there are six terms you can't make a one-to-one associating of particles with potential energy. Second, if someone tried to associate potential energy with a single particle, then that person might mistakenly include too many terms in their sum:
$$U_\text{g}\ne-G\left( \underbrace{\frac{m_1m_2}{d_{1,2}}}_\text{a term for particle 1} + \underbrace{\frac{m_2m_1}{d_{2,1}}}_\text{a term for particle 2} + \cdots \right)$$ This incorrect expression would over-count the number of terms in $U_\text{g}$.
To avoid such over-counting, and to make our language more inline with the mathematics, we tend to speak of the potential energy of the system.
$^\text{1}$The expression $-\frac{Gm_1m_2}{d_{1,2}}$ is a more general form of the more-common $U=mgh$, but that's a separate issue.
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