After submitting to correct some of your very confused notations especially your incorrect concept of independence of probabilities about observed different data samples instead of the correct concept of independence of different random vectors (also see @whuber's question commented under your own latest confused answer), the linked post in the comment hasn't completely clarified this question for you.

Please bear in mind that here $(X_i,Y_i)$ is a single (atomic) random vector while your concerned question is not about the usual conditional independence of random vectors which should be expressed as $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j) | Z, \forall i \neq j$ following your notation. Yet your real question is about independence of the split conditional random variables $Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$ which is not directly provable by plugging to any of the referenced theorem which basically says if $X \mathrel{\unicode{x2AEB}} Y$ then $g(X) \mathrel{\unicode{x2AEB}} h(Y), \forall g, h$. Before trying that out indirectly we can first use a simple example to feel this subtle kind of 'split-conditional' pairwise independence.

Suppose your DGP follows an additional simple deterministic rule $Y_i=X_i, \forall i$, while all the random vectors are still pairwise independent $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j), \forall i \neq j$. In this case apparently the conditional random variable $Y_i|X_i=X_i$ trivially, and note the $X$-component projection of each vector $Z_i=(X_i,Y_i)$ is a function of random vectors. Then we can apply above referenced theorem to rigorously prove $X_i \mathrel{\unicode{x2AEB}} |X_j, \forall i \neq j$, so we have $Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$.

Now to be able to rigorously prove the general case, let $l_i$ be above projection function onto the $X$ component of each random vector $Z_i=(X_i,Y_i)$. Then a generic conditional random variable $Y_i|X_i \sim p_i(X_i)=p_i(l_i(Z_i))$ which is still a (composite) function of the random vector $Z_i$ (here $p_i$ is a general distribution function assuming each conditional random variable $Y_i|X_i$ is well-defined), therefore we now can prove the same independence conclusion in the most general case only assuming independence of the sequence of random vectors.

After submitting to correct some of your very confused notations especially your incorrect concept of independence of probabilities about observed different data samples instead of the correct concept of independence of different random vectors (also see @whuber's question commented under your own latest confused answer), the linked post in the comment hasn't completely clarified this question for you.

Please bear in mind that here $(X_i,Y_i)$ is a single (atomic) random vector while your concerned question is not about the usual conditional independence of random vectors which should be expressed as $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j) | Z, \forall i \neq j$ following your notation. Yet your real question is about independence of the split conditional random variables $Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$ which is not directly provable by plugging to any of the referenced theorem which basically says if $X \mathrel{\unicode{x2AEB}} Y$ then $g(X) \mathrel{\unicode{x2AEB}} h(Y), \forall g, h$. Before trying that out indirectly we can first use a simple example to feel this subtle kind of 'split-conditional' pairwise independence.

Suppose your DGP follows an additional simple deterministic rule $Y_i=X_i, \forall i$, while all the random vectors are still pairwise independent $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j), \forall i \neq j$. In this case apparently the conditional random variable $Y_i|X_i=X_i$ trivially, and note the $X$-component projection of each vector $Z_i=(X_i,Y_i)$ is a function of random vectors. Then we can apply above referenced theorem to rigorously prove $X_i \mathrel{\unicode{x2AEB}} |X_j, \forall i \neq j$, so we have $Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$.

Now to be able to rigorously prove the general case, let $l_i$ be above projection function onto the $X$ component of each random vector $Z_i=(X_i,Y_i)$. Then a generic conditional random variable $Y_i|X_i \sim p_i(X_i)=p_i(l_i(Z_i))$ which is still a (composite) function of the random vector $Z_i$ (here $p_i$ is a general distribution function assuming each conditional random variable $Y_i|X_i$ is well-defined), therefore we now can prove the same independence conclusion in the most general case only assuming independence of the sequence of random vectors.

Finally back to the common regression cases as you've already understood a fixed identical conditional distribution $p_{Y|X}$ is assumed implicitly for each conditional rv $Y_i|X_i$, we can still assert they're independent even if their conditional distribution functions are the same as an extreme edge case because any value $y_i$ of a conditional rv cannot give any clue about another conditional rv's value $y_j$ assuming IID data for regression.

After submitting to correct some of your confused notations especially your incorrect concept of independence of probabilities about observed different data samples instead of the correct concept of independence of different random vectors, it's not quite true the linked post in the comment under your question can clarify your question as you self-claimed several months ago.

Please bear in mind that here $(X_i,Y_i)$ is a single (atomic) random vector while your concerned question is not about the usual conditional independence of random vectors which should be expressed as $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j) | Z, \forall i \neq j$ following your notation and can be easily verified from my linked Wikipedia reference. Yet your real question is about independence of the splitted conditional random vectors $X_i|Y_i \mathrel{\unicode{x2AEB}} X_j|Y_j, \forall i \neq j$ which unfortunately may not be necessarily true. For example, your DGP can follow the rule $X_i=Y_i$ for every random vector, while all the random vectors are still pairwise independent, i.e., $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j), \forall i \neq j$.

After submitting to correct some of your very confused notations especially your incorrect concept of independence of probabilities about observed different data samples instead of the correct concept of independence of different random vectors (also see @whuber's question commented under your own latest confused answer), the linked post in the comment hasn't completely clarified this question for you.

Please bear in mind that here $(X_i,Y_i)$ is a single (atomic) random vector while your concerned question is not about the usual conditional independence of random vectors which should be expressed as $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j) | Z, \forall i \neq j$ following your notation. Yet your real question is about independence of the split conditional random variables $Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$ which is not directly provable by plugging to any of the referenced theorem which basically says if $X \mathrel{\unicode{x2AEB}} Y$ then $g(X) \mathrel{\unicode{x2AEB}} h(Y), \forall g, h$. Before trying that out indirectly we can first use a simple example to feel this subtle kind of 'split-conditional' pairwise independence.

Suppose your DGP follows an additional simple deterministic rule $Y_i=X_i, \forall i$, while all the random vectors are still pairwise independent $(X_i,Y_i) \mathrel{\unicode{x2AEB}} (X_j,Y_j), \forall i \neq j$. In this case apparently the conditional random variable $Y_i|X_i=X_i$ trivially, and note the $X$-component projection of each vector $Z_i=(X_i,Y_i)$ is a function of random vectors. Then we can apply above referenced theorem to rigorously prove $X_i \mathrel{\unicode{x2AEB}} |X_j, \forall i \neq j$, so we have $Y_i|X_i \mathrel{\unicode{x2AEB}} Y_j|X_j, \forall i \neq j$.

Now to be able to rigorously prove the general case, let $l_i$ be above projection function onto the $X$ component of each random vector $Z_i=(X_i,Y_i)$. Then a generic conditional random variable $Y_i|X_i \sim p_i(X_i)=p_i(l_i(Z_i))$ which is still a (composite) function of the random vector $Z_i$ (here $p_i$ is a general distribution function assuming each conditional random variable $Y_i|X_i$ is well-defined), therefore we now can prove the same independence conclusion in the most general case only assuming independence of the sequence of random vectors.