Prove or disprove that the sum of Markov processes is Markov 
Suppose $X$ and $Y$ are Markov Processes, is their sum a Markov process?
$\require{color}$

Intuition tells me that in general this should not hold. I want to verify this mathematically.
I am aware of specific counterexamples of the above question; however, I am interested in knowing wether the below approach is viable:
For simplicity consider the setting of $(X_k)_{k=0}^n$ and $(Y_k)_{k=0}^n$ being discrete-time Markov Chain (I believe the argument is easily extended to other cases as well), and define $Z_k := X_k+Y_k$, $k\geq 0$. Then
$$\begin{aligned}
P(Z_{n}=z_{n}\vert Z_{n-1}=z_{n-1},\dots,Z_{0}=z_{0})&=P(X_{n}+Y_{n}=z_{n}\vert X_{n-1}+Y_{n-1}=z_{n-1},\dots,X_{0}+Y_{0}=z_{0})\\
&=\sum_{y_0,\dots,y_n}P(X_{n}+Y_{n}=z_{n}\vert X_{n-1}+Y_{n-1}=z_{n-1},\dots,X_{0}+Y_{0}=z_{0},{\color{red}{Y_n=y_n,\dots,Y_0=y_0}})P(Y_n=y_n,\dots,Y_0=y_0)\\
&=\sum_{y_0,\dots,y_n}P(X_{n}=\underbrace{z_{n}-y_{n}}_{\text{not a R.V.}}\vert X_{n-1}=z_{n-1}-y_{n-1},\dots,X_{0}=z_{0}-y_0)P(Y_n=y_n,\dots,Y_0=y_0)\\
&{\color{blue}{=}}\sum_{y_0,\dots,y_n}P(X_{n}=z_{n}-y_{n}\vert X_{n-1}=z_{n-1}-y_{n-1})P(Y_n=y_n,\dots,Y_0=y_0)\\
&=\sum_{y_0,\dots,y_n}P(X_{n}=z_{n}-y_{n}\vert X_{n-1}=z_{n-1}-y_{n-1})P(Y_n=y_n\vert Y_{n-1}=y_{n-1})P(Y_{n-1}=y_{n-1},\dots,Y_0=y_0).
\end{aligned}$$
would this be going in the right direction? From this last line you cannot conclude much without any further assumption on the relationship between $X$ and $Y$, which would make the implication in the question false for general $X$ and $Y$.
In $({\color{blue}{=}})$ I used the Markov Property of $X$ and in the last equality that of $Y$.

Edit:
Can the last sum be reduced/written in the form
$$P(Z_{n}=z_{n}\vert Z_{n-1}=z_{n-1}){\color{red}\cdot(\text{some other terms})}$$
so that it would be clear that the Markov Property would not hold in general with the given assumptions?
 A: Given that you are not looking for a specific counter-example, but perhaps for a fundamental reason on why it fails, here are my few cents on the problem.
Intuition. If you know the value of the macroscopic variable $X(t)+Y(t)$ at the present time $t$, you cannot tell how it will evolve next without zooming into the microscopic state $\left(X(t),Y(t)\right)$. This is fundamentally what breaks the Markovianity down. Imagine that you have a collection of $N$ distinct binary Markov processes (or chains) $X_i(t)\in\left\{0,1\right\}$ for $i=1,2,\ldots,N$. In this slightly simpler framework, your question could be cast as: is the fraction of processes at state $1$ a Markov process? I.e., is the macroscopic state-variable $Z(t)\overset{\Delta}= \frac{1}{N}\sum_{i=1}^N X_i(t)$ Markov? No, because if you know that $30\%$ of the processes are at state '1', you cannot tell what is the rate of transitioning to, e.g., $0.3+1/N$ without having to look at the identity of the processes at state '1', i.e., without looking at the microscopic state-variable $\left(X_1(t),X_2(t),\ldots,X_N(t)\right)$. Unless, of course, in some degenerated framework where the $X_i's$ have the same dynamics (or distribution) and they are all independent.
Mathematically. Let $f(x,y,z)\overset{\Delta}=\mathbb{P}\left(X(t)+Y(t)=z\left|X(t)=x,Y(t)=y\right.\right)$. In the framework where the distribution of $X$ and $Y$ are distinct, then $f(x,y,z)\neq f(y,x,z)$, i.e., $f$ is not symmetric in the first two variables (or exchangeable in the higher-dimensional framework counterpart). On one hand, we have
$$\mathbb{P}\left(X(t+1)+Y(t+1)=n\left|X(t)+Y(t)=m\right.\right)=\sum_{w\in \mathcal{E}}\mathbb{P}\left(X(t+1)+Y(t+1)=n\left|X(t)=m-w,Y(t)=w\right.\right)\mathbb{P}\left(Y(t)=w\right)=\sum_{w\in \mathcal{E}} f(m-w,w,n)\mathbb{P}\left(Y(t)=w\right)$$,
and on the other hand, we have
$$\mathbb{P}\left(X(t+1)+Y(t+1)=n\left|X(t)+Y(t)=m\right.\right)=\sum_{w\in \mathcal{E}}\mathbb{P}\left(X(t+1)+Y(t+1)=n\left|X(t)=w,Y(t)=m-w\right.\right)\mathbb{P}\left(X(t)=w\right)=\sum_{w\in \mathcal{E}} f(w,m-w,n)\mathbb{P}\left(X(t)=w\right)$$.
Only by accident (or in some degenerated frameworks) these two coincide and from here you can build a family of counter-examples.
In summary, $X+Y$ is not Markov because the function $g(n,m)=\mathbb{P}\left(X(t+1)+Y(t+1)=n\left|X(t)+Y(t)=m\right.\right)$ is not well-defined in the first place, in general.
