Suppose $\nu$ is a finite measure on $X$ and the transition kernel $P$ is given. Then, by the Lebesgue decomposition, we have $$P(x,A) = \int_A p(x,y) \nu(dy) + m(x)1_A (x)$$where $p(x,y)\ge 0$, and $m(x)$ is allowed to be non-zero only if $\nu(\{x\})=0$, and $$m(x) + \int_X p(x,y) \nu(dy)=1.$$
How do we get such a Lebesgue decomposition with a function $m(x)$ with the specific conditions given above? From the Lebesgue decomposition, we should have $$P(x,A) = \int_A p(x,y)\nu(dy) + P_{\perp}(x,A)$$ where $p(x,y)$ is the density of $P(x,\cdot)$ with respect to $\nu$ and $P_{\perp}$ is orthogonal to $\nu$. However, $m$ here only depends on $x$ and I can't see how to obtain such a representation from the Lebesgue decomposition.
Now given these assumptions, we have the following propositon.
I can follow the proof to get (4.13). However, from here, how do we get (4.11)? Using (4.12), I can see that the first and third terms combine to give $\int_A p^n(x,y)\nu(dy)$, but the second term survives. From the usual Lebesgue decomposition, $m(z)$ should be singular with respect to $\nu$, so the multiplication with the density with respect to $\nu$ should vanish. But how do we get this result with only the assumption that $m(x)$ is non-zero provided $\nu(x)=0$?
Finally, I am not sure we actually get (4.12). For instance, if we take $k=0, n=1$, then we get $p(x,y)=P(x,X)=1$, unless we have a particular definition for $p^0(z,y)$. How can we actually prove this equality?
I agree that the definition of $p^n$ seems to not satisfy the claimed identity, so this is only an answer to the first part of your question:
There must be some other assumption on the interplay between $\nu$ and $P$ in order for the equation to hold. It is not enough that $P$ is a transition kernel and $\nu$ a finite measure. Take, for instance, $X=\mathbb{R}$ and $\nu=\delta_0$ and $P(x,A)=\lambda(A\cap [x-2,x-1]),$ where $\lambda$ denotes the Lesbegue Measure. This is clearly a measure in $A$ and measurable in $x$, hence a valid kernel.
Then, observing $P(0,A),$ it's a probability measure supported on $[-1,-2]$. However, for any $p(0,y)\geq 0$ and any $m\geq 0$, we have for $A=[-1,-2]$ that
$$ \int_A p(0,y)\textrm{d} \delta_0+m(0)1_A(0)=0, $$ so this decomposition can never give $P(0,A)$.