Total probability of Conditional Probability Suppose that we have a conditional probability $P(x | y)$ and a partition $x'$ of the sample space. How can we apply the law of total probability to conclude that
$$P(x|y) = \sum\limits_{x'}^{} P((x|y)\cap x')?$$
 A: It is rather straightforward, use the fact that for any events $A$ and $B$
$$P( A \cap B )= P(B)P(A|B).$$
Thus $P((x|y) \cap x^{\prime}) = P((x|y)|x^{\prime})P(x^{\prime})$, so by the law of total probability
$$P(x|y) = \sum_{x^{\prime} }P((x|y)|x^{\prime})P(x^{\prime})=\sum_{x^{\prime} } P((x|y) \cap x^{\prime}).$$
Caution: Using $x^{\prime}$ as the elements from the partition is not the best notation, $\prime$ is often used to denote the complement of an event. 
A: In addition, I think this answer also provides a good view of what happens here.
Proof of Total Probability Theorem for Conditional Probability
I find it rather confusing to use condition with 
$$\sum_{z}P((x|y)|z)P(z) \\\text{z here in place of x' in question}$$
for the event space, $\Omega$ for $Z$ here is not so clear. The space for $Z$, or the value of $P(Z)$ here should be defined over $\Omega_y \cap \Omega_Z$.
This makes the overall answer be either:
$$P(x|y) = \sum^{\Omega_y\cap \Omega_Z}_zP(x|y\cap z)P(z) = \sum_zP(x\cap z|y)$$
Also, as mentioned by Graham Kemp in the comment:

∣ is not a set operation; it is the divider between the event being measured and the condition under which the measured. There can only be at most one such divider in a probability measure function.

