Let $M, N$ be two connected closed orientable smooth $n$-manifolds, where $n\geq 3$. Consider the connected sum $M\sharp N:=\frac{\big(M\backslash\varphi(\Bbb B^n)\big)\sqcup \big(N\backslash\psi(\Bbb B^n)\big)}{\sim}$ where $\varphi\colon \overline{\Bbb B^n}\hookrightarrow M$ is an orientation-preserving, $\psi\colon \overline{\Bbb B^n}\hookrightarrow M$ is orientation-reversing smooth embedding and $\sim$ is defined by identifying $\psi(z)$ with $\varphi(z)$ for all $z\in \Bbb S^{n-1}$.
So, $M\sharp N$ is closed orientable smooth $n$-manifold and $\mathcal S:=\big\{p\in M\sharp N: p=\big[\varphi(z)\big]=\big[\psi(z)\big] \text{ for some }z\in \Bbb S^{n-1}\big\}$ is a smoothly embedded $(n-1)$-sphere in $M\sharp N$.
Attach an $n$-ball $\overline{\Bbb B^n}$ to $M\sharp N$ using $\mathcal S$, and consider the inclusion map $i\colon M\sharp N\hookrightarrow (M\sharp N)\amalg \overline{\Bbb B^n}$. Crushing the image of $\overline{\Bbb B^n}$ inside $(M\sharp N)\amalg \overline{\Bbb B^n}$ we have a homotopy equivalence $f\colon (M\sharp N)\amalg \overline{\Bbb B^n}\to M\lor N$.
Let $[M]$ and $[N]$ be two generators of $H_n(M)$ and $H_n(N)$, respectively. Here, coefficient ring is $\Bbb Z$.
Is it possible to choose a generator $[M\sharp N]\in H_n(M\sharp N)$ so that $[M\sharp N]$ goes to $[M]\oplus [N]\in H_n(M)\oplus H_n(N)$ under the composition of maps $$H_n(M\sharp N)\xrightarrow{i_*} H_n\big((M\sharp N)\amalg \overline{\Bbb B^n}\big)\xrightarrow{f_*} H_n(M\lor N)\cong H_n(M)\oplus H_n(N)?$$
I believe that one has to observe how the fundamental class of $M\sharp N$ looks like in terms of $[M]$ and $[N]$ giving some simplicial structures on each $M, N$. If not is there any other process?
Yes, this is possible. To do so, I will first give an explicit description of the fundamental class $[M\#N]$ in terms of the fundamental classes $[M\setminus\varphi(B^n)]$ and $[N\setminus\psi(B^n)]$, which is an illustration of a general principle, and then compare the fundamental classes $[M\setminus\varphi(B^n)]$ and $[N\setminus\psi(B^n)]$ with the fundamental classes $[M]$ and $[N]$ respectivly.
Let $i_1\colon M\setminus\varphi(B^n)\rightarrow M\#N$ and $i_2\colon N\setminus\psi(B^n)\rightarrow M\#N$ be the respective canonical inclusions. Consider the following commutative diagram $$\require{AMScd} \begin{CD} 0=H_n(S) @>>> H_n(M\#N) @>>> H_n(M\#N,S) @>{\partial}>> H_{n-1}(S)\\ @. @. @A{(i_1)_{\ast}\oplus(i_2)_{\ast}}AA @AAA \\ @. @. H_n(M\setminus\varphi(B^n),\varphi(S^{n-1}))\oplus H_n(N\setminus\psi(B^n),\psi(S^{n-1})) @>{\partial\oplus\partial}>> H_{n-1}(\varphi(S^{n-1}))\oplus H_{n-1}(\psi(S^{n-1})) \end{CD}.$$ Using collar neighborhoods and Mayer-Vietoris, we see that the vertical map $(i_1)_{\ast}\oplus(i_2)_{\ast}\colon H_n(M\setminus\varphi(B^n),\varphi(S^{n-1}))\oplus H_n(N\setminus\psi(B^n),\psi(S^{n-1}))\rightarrow H_n(M\#N,S)$ is an isomorphism. The horizonal maps come from the LES of pairs respectively. Note that the lower horizontal map maps $([M\setminus\varphi(B^n)],[N\setminus\psi(B^n)])\mapsto ([\varphi(S^{n-1})],[\psi(S^{n-1})])$, which the vertical map maps to $0$, since $\varphi$ and $\psi$ are oppositely oriented. From exactness of the top row, it follows that $(i_1)_{\ast}[M\setminus\varphi(B^n)]+(i_2)_{\ast}[N\setminus\psi(B^n)]$ is the image of a unique element $[M\#N]\in H_n(M\#N)$ under the map $H_n(M\#N)\rightarrow H_n(M\#N,S)$. This element is, in fact, the fundamental class of $M\#N$ with the given orientation, but since this is not necessary to know in the following argument, I will leave that as an exercise.
Everything in the previous paragraph works verbatim when you glue any two manifolds along an orientation-reversing homeomorphism of their boundaries.
Next, consider the quotient maps $r_1\colon M\setminus\varphi(B^n)\rightarrow(M\setminus\varphi(B^n))/\varphi(S^{n-1})\cong M$ and $r_2\colon N\setminus\varphi(B^n)\rightarrow(N\setminus\psi(B^n))/\psi(S^{n-1})\cong N$. These homeomorphisms come from noting that the embeddings of $\overline{B^n}$ into $M$ and $N$ respectively extend to embeddings of $2B^n$ (the ball of radius $2$, to be clear) into $M$ and $N$ respectively and that $(2B^n\setminus B^n)/S^{n-1}\cong 2B^n$ by a homeomorphism that is the identity outside of compact subspace. Then, $r_1$ and $r_2$ induce isomorphisms $H_n(M\setminus\varphi(B^n),\varphi(S^{n-1})\rightarrow H_n(M)$ and $H_n(N\setminus\psi(B^n),\psi(S^{n-1}))\rightarrow H_n(N)$ respectively. Note also that $r_1$ and $r_2$ restrict to homeomorphisms between $\mathrm{int}(M\setminus\varphi(B^n))$ and $\mathrm{int}(M)$, resp. $\mathrm{int}(N\setminus\psi(B^n))$ and $\mathrm{int}(N)$. For each $x\in\mathrm{int}(M\setminus\varphi(B^n))$, we obtain a commutative diagram $$\require{AMScd} \begin{CD} H_n(M\setminus\varphi(B^n),\varphi(S^{n-1})) @>>> H_n(M\setminus\varphi(B^n),M\setminus\varphi(B^n)-x)\\ @VVV @VVV \\ H_n(M) @>>> H_n(M,M-r_1(x)) \end{CD}. $$ These maps are all isomorphisms. This implies that $(r_1)_{\ast}[M\setminus\varphi(B^n)]=[M]$ and analogously $(r_2)_{\ast}[N\setminus\psi(B^n)]=[N]$. I recommend thinking through a simple example, such as $S^2\#S^2$, visually to get a feeling for this being exactly the behavior we expect geometrically.
Now, for the finishing touch, we just have to compare some diagrams. There is a map $r\colon M\#N\rightarrow M\lor N$, obtained by collapsing $S$ (this is the same thing as first attaching a $B^n$ along $S$ and then collapsing that, as in your post). There are also maps $p_1\colon M\lor N\rightarrow M$ and $p_2\colon M\lor N\rightarrow N$, obtained by collapsing the respective other summand to the wedge point. Note that the sum of the maps induced by $p_1$ and $p_2$ furnishes the standard isomorphism $H_n(M\lor N)\rightarrow H_n(M)\oplus H_n(N)$. These maps fit into the following commutative diagrams $$\require{AMScd} \begin{CD} M\#N @>{r}>> M\lor N @. M\#N @>{r}>> M\lor N\\ @A{i_1}AA @V{p_1}VV @A{i_2}AA @V{p_2}VV\\ M\setminus\varphi(B^n) @>{r_1}>> M @. N\setminus\psi(B^n) @>{r_2}>> N \end{CD}. $$ On homology, we obtain the following commutative diagram $$\require{AMScd} \begin{CD} H_n(M\#N) @= H_n(M\#N) @>{r_{\ast}}>> H_n(M\lor N)\\ @VVV @. @V{((p_1)_{\ast},(p_2)_{\ast})}VV\\ H_n(M\#N,S) @<{(i_1)_{\ast}\oplus(i_2)_{\ast}}<< H_n(M\setminus\varphi(B^n),\varphi(S^{n-1}))\oplus H_n(N\setminus\psi(B^n),\psi(S^{n-1})) @>{(r_1)_{\ast}\oplus(r_2)_{\ast}}>> H_n(M)\oplus H_n(N) \end{CD}. $$ The composite $H_n(M\#N)\rightarrow H_n(M\lor N)\cong H_n(M)\oplus H_n(N)$ maps $[M\#N]\mapsto([M],[N])$ by commutativity of the diagram and the previous considerations, as desired and expected.