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Let

• 
  $(Omega,mathcal A,operatorname P)$ be a probability space


• 
  $(mathcal F_t)_{tge0}$ be a filtration on $(Omega,mathcal A)$
  


• 
  $E$ be a Polish space and $mathcal E:=mathcal B(E)$
  


• 
  $(X_t)_{tge0}$ be an $E$-valued $mathcal F$-Markov process on $(Omega,mathcal A,operatorname P)$
  


• 
  $kappa_{s,:t}$ be a regular version of the conditional probability of $X_t$ given $X_s$, i.e. $kappa_{s,:t}$ is a Markov kernel on $(E,mathcal E)$ with $$operatorname Pleft[X_tin Bmid X_sright]=kappa_{s,:t}(X_s,B);;;text{almost surely for all }Binmathcal Etag1$$ for $s,tge0$
  

By the Markov property and $(1)$, $$operatorname Pleft[X_tin Bmidmathcal F_sright]=kappa_{s,:t}(X_s,B);;;text{almost surely for all }Binmathcal Etext{ and }0le sle t.tag2$$

Usually, we want $(kappa_{s,:t}:0le sle t)$ to satisfy the Chapman-Kolmogorov equation $$kappa_{r,:t}=kappa_{r,:s}kappa_{s,:t},tag3$$ where the right-hand side denotes the composition of transiton kernels, for all $0le rle sle t$. However, with the definition of $kappa_{s,:t}$ as the conditional probability of $X_t$ given $X_s$, we've only got$^1$ $$kappa_{r,:t}(x,B)=(kappa_{r,:s}kappa_{s,:t})(x,B);;;text{for all }Binmathcal Etext{ and }operatorname Pcirc:X_r^{-1}text{-almost all }xin Etag4$$ for all $0le sle t$. However, in the literature, one is usually assuming that $(2)$ and $(3)$ hold together. Why is that possible?

I could imagine that the reason is the following: Since $E$ is Polish, given $(kappa_{s,:t}:0le sle t)$ with $(3)$ there is always a Markov process $tilde X$ on an other probability space with transition semigroup $(kappa_{s,:t}:0le sle t)$ and initial distribution $operatorname Pcirc:X_0^{-1}$. Is that the correct argument?

$^1$ Note that there is a crucial selection of a common null set for all $B$ happening in $(4)$. I guess this is legitimate as long as $mathcal E$ is countably generated. Maybe someone could comment on this.

My (perhaps naive) interpretation of your question is that it deals with "surely" versus "almost surely" issues. For fixed times $r < s < t$, the following equation only holds in an “almost sure” sense:
$$ k_{r,t} = k_{r,s} cdot k_{s,t} quad (Eq. 1)$$
To see this, suppose it holds surely (for all corresponding points $x$ and sets $B$ for which we can evaluate $k_{r,t}(x,B)$). Let $tilde{k}_{r,t}$ be another “version” of $k_{r,t}$ that is different from $k_{r,t}$ on a set of measure 0. Then we want to also have
$$tilde{k}_{r,t} = k_{r,s} cdot k_{s,t} quad (Eq. 2)$$ surely, but this is impossible since we have changed the left-hand-side of equation (1) without changing the right-hand-side. The only way it makes sense is that both equations (1) and (2) are only guaranteed to hold in an almost-sure sense.

An example way to change $k_{r,t}$ to a different version $tilde{k}_{r,t}$ is this: Suppose we can identify an $x^*$ such that $P[X_r=x^*]=0$. Then define for all corresponding $x, B$:
$$ tilde{k}_{r,t}(x,B) = left{ begin{array}{ll}
k_{r,t}(x,B) &mbox{ if $x neq x^*$} \
mu(B) & mbox{ if $x =x^*$}
end{array}
right.$$
where $mu(B)$ is some measure over events $B$ that is different from $k_{r,t}(x^*,B)$.