While defining tensor product $M \otimes N$, we take the submodule $D$ of the free $A$-module $C = A^{(M \times N)}$ where $D$ is generated by elements of the form $(x+x',y)-(x,y)-(x',y), (ax,y)-a(x,y)$ etc with $x,x' \in M, y \in N$ and define $T = C/D$.
Here can we use componentwise addition,multiplication and say $(x+x',y)-(x,y)-(x',y) = (x',0)-(x',y) = (0,-y)$ and $(ax,y)-a(x,y) = (ax,y) - (ax,ay) = (0,y-ay)$, thus D being generated by elements of the form $(0,-y), (0,y-ay), (-x,0), (x-ax,0)$?
Edit: I don't think we're allowed such componentwise addition,multiplication as otherwise in $T = C/D$, we'd have $a \otimes b + c \otimes d = ((a,b)+D) + ((c,d)+D) = ((a,b)+(c,d)) + D = (a+c,b+d) + D = (a+c)\otimes(b+d)$.
This certainly must not be true. But I can't understand why.
The elements of the $A$-module $A^{(M\times N)}$ are of the form $\sum_{i=1}^n a_i(x_i, y_i)$ where $a_i \in A$, $x_i \in M$, and $y_i \in N$. We define addition by
$$\sum_{i=1}^n a_i(x_i, y_i) + \sum_{i=1}^n b_i(x_i, y_i) = \sum_{i=1}^n (a_i + b_i)(x_i, y_i)$$
and $A$-multiplication by
$$c\left(\sum_{i=1}^n a_i(x_i, y_i)\right) = \sum_{i=1}^n (c_{}\,a_i)(x_i, y_i).$$
According to these rules, you cannot simplify $(x + x',y) - (x, y) - (x', y)$ or $(ax, y) - a(x, y)$ in the ways that you did.
In $M\oplus N$ however, we have $(x, y) + (x', y') = (x + x', y + y')$ and $c(x, y) = (cx, cy)$. These were the rules you were using to make the simplifications. This is where I believe your confusion lies.