Related question: The sum of two elementary processes is an elementary process
We work with respect to a filtered probability space $(\Omega,\mathcal F,\{\mathcal{F_t\}_{t\ge 0},P})$
An elementary process is of the form
\begin{equation} \label{eq:1} \xi(t) = Z_01_{\{t=0\}}+\sum_{k=1}^n Z_k1_{\{s_k<t\le t_k\}} \end{equation} for ${n\ge 0}$, times $0 \le s_1 < t_1 \le s_2 < t_2 \le \dots \le s_n < t_n \,$, ${\mathcal{F}_0}$-measurable random variable ${Z_0}$ and ${\mathcal{F}_{s_k}}$-measurable random variables ${Z_k}$.
The integral of an elementary process with regard to a stochastic process $X$ over a finite range $[0,t]$ is defined by
\begin{equation} \int_0^t\xi\,dX \equiv \sum_{k=1}^nZ_k(X(t_k\wedge t)-X(s_k\wedge t)) \end{equation}
If we take
$$\xi_1(t) = Z_0 1_{\{t=0\}} +\sum\limits_{k=1}^{n_1} Z_k 1_{\{s_k < t \le t_k\}}$$ and $$\xi_2(t) = Y_01_{\{t=0\}}+\sum_{k=1}^{n_2} Y_k1_{\{\tilde{s}_k<t\le \tilde{t}_k\}}$$
two elementary processes, how can we show that
$$\int_0^t \xi_1 + \xi_2\,dX =\int_0^t \xi_1 \,dX + \int_0^t \xi_2\,dX$$
As is shown in the accepted answer to your linked question, we can find a sequence of times $0 \leq s_1 < t_1 \leq \dots \leq s_n < t_n$ such that we can represent $\xi_1$ and $\xi_2$ with respect to these times. This kind of refinement of the partition is usually the key when trying to prove properties about integrals involving more than one simple function.
So, without loss of generality, \begin{align*} \xi_1(t) =& Z_0 1_{\{t=0\}} +\sum\limits_{k=1}^{n} Z_k 1_{\{s_k < t \le t_k\}}, \\ \xi_2(t) =& Y_01_{\{t=0\}}+\sum_{k=1}^{n} Y_k1_{\{s_k<t\le t_k\}}. \end{align*} In particular, we have $$\xi_1(t) + \xi_2(t) = (Z_0+Y_0) 1_{\{t=0\}} +\sum\limits_{k=1}^{n} (Z_k+Y_k) 1_{\{s_k < t \le t_k\}}.$$ So we have \begin{align*} \int_0^t \xi_1 + \xi_2\,dX =& \sum_{k=1}^n (Z_k+Y_k)(X(t_k\wedge t)-X(s_k\wedge t))\\=& \sum_{k=1}^n Z_k(X(t_k\wedge t)-X(s_k\wedge t)) + \sum_{k=1}^n Y_k(X(t_k\wedge t)-X(s_k\wedge t)) \\=& \int_0^t \xi_1 \,dX + \int_0^t \xi_2\,dX \end{align*} as desired.