Derivation of the ARMA model as acombination of the AR and MA models On the Wikipedia article on the ARMA model, its derivation is simplified as a combination of the AR and MA models:
AR

$$
X_t = c + \sum_{i=1}^p \varphi_i X_{t-i} + \varepsilon_t
$$

MA

$$
X_t = \mu + \varepsilon_t + \sum_{i=1}^q \theta_i \varepsilon_{t-i}
$$

ARMA

$$
X_t = c + \varepsilon_t + \sum_{i=1}^p \varphi_i X_{t-i} + \sum_{i=1}^q \theta_i \varepsilon_{t-i}
$$

Sum of AR and MA

$$
2X_t = \mu + c + 2\varepsilon_t + \sum_{i=1}^p \varphi_i X_{t-i} + \sum_{i=1}^q \theta_i \varepsilon_{t-i}
$$

At first glance it seems almost like they are simply summed together to form this model, however, $\mu$ is gone, and $\varepsilon_t$ and $X_t$ should be doubled. Clearly this is not an entirely correct interpretation, but it's still close.
Is there a way to explain this discrepancy between the sum of AR and MA and the ARMA model, or is there another more natural way of deriving this model?
 A: No, there is no way to explain this discrepancy because it does not make sense to sum together an AR and MA model in the way you have, because $X_t$ cannot be both an AR model and an MA model at the same time. If you write down 
$$
X_t = \mu + \varepsilon_t + \sum_{i=1}^q \theta_i \varepsilon_{t-i}
$$
with $q$ finite, then you cannot also write down 
$$
X_t = c + \sum_{i=1}^p \varphi_i X_{t-i} + \varepsilon_t,
$$
with $p$ finite, because it is a contradiction. You can call one of these processes $Y_t$, and then write the  sum as $W_t$, however. 

...or is there another more natural way of deriving this model?

Yes, both of these models are types of linear processes. 
A: It's much easier to do this "derivation" with lag operator:
$$Lz_t=z_{t-1}$$ and 
$$\phi(L)=1+\sum_{i=1}^q\phi_iL^i$$
$$\theta(L)=1-\sum_{i=1}^q\theta_iL^i$$
This way MA(q) is $$x_t=\phi(L)\varepsilon_t=c+\varepsilon_t+\sum_{i=1}^q\phi_i\varepsilon_{t-i}$$ and AR(p) is $$\theta(L)x_t=x_t-\sum_{i=1}^p\theta_ix_{t-i}=c$$. You can combine them as you wish. For instance, ARMA(p,q) is $$\theta(L)x_t=c+\phi(L)\varepsilon_t$$
Let's see how ARMA(1,2) works out:
$$x_t-\theta_1x_{t-1}=c+\varepsilon_t+\phi_1\varepsilon_{t-1}+\phi_2\varepsilon_{t-2}$$
$$x_t=c+\theta_1x_{t-1}+\varepsilon_t+\phi_1\varepsilon_{t-1}+\phi_2\varepsilon_{t-2}$$
Note, that in MA process, the mean is equal to a constant:
$$E[x_t]=c$$
In AR or ARMA this is not true:
$$E[x_t]\ne c$$
That is why sometimes the connstant in MA process is denoted with $\mu$ to allude to the common symbol for mean in statistics.
