Suppose $(X, \Sigma_X)$ and $(Y, \Sigma_Y)$ are measurable spaces such that $\Sigma_Y$ is generated by a set $B$.
Suppose $k : X \times \Sigma_Y \to [0, 1]$ has the property that $k(x, -)$ is a (sub-)probability measure for each $x \in X$, and $k(-, E)$ is measurable for every $E \in B$, wrt the Borel $\sigma$-algebra on $[0, 1]$.
Does that make $k(-, U)$ measurable for each $U \in \Sigma_Y$? (in other words, is $k$ a Markov kernel?)
$\newcommand{\N}{\mathbb N} \newcommand{\R}{\mathbb R} \newcommand{\B}{\mathcal B} \newcommand{\F}{\mathcal F} \newcommand{\X}{\mathcal X} \newcommand{\ep}{\epsilon} \newcommand{\la}{\lambda} \newcommand{\si}{\sigma} \newcommand{\Si}{\Sigma} \renewcommand{\c}{\circ} \newcommand{\tr}{\operatorname{tr}} \newcommand{\E}{\operatorname{\mathsf E}}$
The answer is no in general. Indeed, suppose $B=\{E,F\}$ and $k(x,A)$ equals $1/2-f(x),f(x),1/2-f(x),f(x)$ if $A$ is $E\setminus F,E\cap F,F\setminus E,Y\setminus E\setminus F$, respectively, where $f\colon X\to[0,1/2]$ is not $\Si_X$-measurable. Then $k(\cdot, U)$ is measurable for each $U\in B$ but not for $U=E\cap F\in \Sigma_Y$.
However, your desired conclusion will hold if $B$ is an algebra. Indeed, suppose that $B$ is an algebra and consider the set $C$ of all $U\in\Si_Y$ such that $k(\cdot, U)$ is measurable. Clearly, $C$ is a monotone class. Hence, your desired conclusion follows by the monotone class theorem.
