Consider the following Proposition from Michael atiyah commutative algebra. I want to show that d is well defined, but I am stuck at that point.
Here we made two choices in order to choose that $y^{\prime}$. In the same way we could have $f(x_2) = u^{\prime}(y^{\prime \prime})$ for some $x_2 \in M$. Then we could deduce $f(x - x_2) = u^{\prime}(y^{\prime} - y^{\prime \prime})$. But why is it true that $y^{\prime} - y^{\prime \prime} \in im(f)$ ?
Let's say we pick $y'' \in u'^{-1}(x_2)$. Like you mentioned, we have that $u'(y'-y'') = f(x-x_2)$. But since $v(x) = v(x_2) = x''$, we have that $x-x_2 \in \ker(v) = \operatorname{im}(u)$. Hence there exists some $x' \in M'$ such that $u(x') = x-x_2$. Now, $u'(f'(x')) = f(x-x_2)$. Hence $$u'(f'(x') - (y'-y'')) = u'(f'(x')) - u'(y'-y'') =f(x-x_2) - f(x-x_2) = 0$$ But $u'$ is injective so $f'(x') = y'-y''$. This means that $y' - y'' \in \operatorname{im}(f')$, hence their projections to the cokernel are identical.