Let $(X, \Sigma_X)$ and $(Y, \Sigma_Y)$ be measurable spaces. (I'm happy to restrict this to standard Borel spaces.)
Suppose there are finite measures $\mu$ and $\nu$ on $X$ and a finite kernel $k$ from $X$ to $Y$ (i.e. $k : X \times \Sigma_Y \to \mathbb{R}$).
Suppose $f, g : X \times Y \to \mathbb{R}$ are such that for all $y \in Y$, $$ \int_X f(x, y) \mu(dx) = \int_X g(x, y)\nu(dx)$$
Does the following equality hold? $$ \int_X \int_Y f(x, y) k(x, dy) \mu(dx) = \int_X \int_Y g(x, y) k(x, dy) \nu(dx)$$
Not true. Let $X=Y=(-1,1)$ with the Borel sigma algebra. Let $k(x,A)=(1+x) \lambda (A)$, (where $\lambda$ is the Lebesgue measure), $\mu=\lambda$ and $\nu =2\mu$. Let $f(x,y)=x$ and $g(x,y)=xy$. Then $\int f(x,y)d\mu (x)=0=\int g(x,y) d\nu (x)$. But the conclusion does not hold.