I don't understand a notation. Consider a measurable space $(S,\mathcal{S})$ and $A\in\mathcal{S}$, a probability kernel $K$ and a RV $X_s$. Does this notation $$\displaystyle \int\limits_SK_{t,u}(y,A)K_{s,t}(X_s,dy)$$
mean that we integrate with respect to $y$ regarding a certain distribution $\mathbb{Q}_x=K_{s,t}(x,\cdot)$ depending on the current value of $X_s$?
It is clear if we have $K_{s,t}(z,dy)$ where $z$ is some fixed valued, but I don't understand what this means when I have a random variable $X_s$ in the kernel.
Forget the fact that $X_s$ is random; assume we have some $x\in S$. Then $K_{s,t}(x,\cdot)$ is a probability measure, and so integration against it makes perfect sense. Now if $\Omega$ is some probability space and $Y:\Omega\to S$ some random variable, then for each $\omega\in\Omega$, we still have that $K_{s,t}(Y(\omega),\cdot)$ is a measure and integration against it can be defined. Explicitly,
$$\int_SF(y)K_{s,t}(Y(\omega),dy)=:I(\omega)$$
is a well-defined function of $\omega\in\Omega$ whenever $F:S\to\mathbb R$ is integrable. This function is measurable since, by definition of a kernel, $x\mapsto K_{s,t}(x,\cdot)$ is measurable. As usual with probability, we suppress the dependence on $\omega\in\Omega$ and just write
$$\int_SF(y)K_{s,t}(Y,dy)=I$$
which is just a random variable. In our specific situation, $\{X_t\}$ is a Markov process with kernel $K$. We know that
$$\int_SK_{t,u}(y,A)K_{s,t}(x,dy)=\mathbb P(X_u\in A|X_s=x)$$
for every $x\in S$ by the Chapman-Kolomogorov equations. In particular, this implies
$$\int_SK_{t,u}(y,A)K_{s,t}(X_s,dy)=\mathbb P(X_u\in A|X_s).$$