Consider the function $f: [0,1] \longrightarrow \mathbb{R}$
$$f(x)= \begin{cases} \frac{m}{n} &\text{ if } x \in \mathbb{Q}, x = \frac{m}{n}, (m,n) = 1 \\ 0 &\text{ if } x \in \mathbb{R} \setminus \mathbb{Q}\end{cases} $$ where $(m,n)$ is the greatest common divisor of $m,n$. This function is NOT Riemann Integrable on $[0,1]$ but I have a problem with the special case where $m = 0$. The author of the question failed to specify that $m \neq 0$, though we have $\gcd(n,m) = 1$.
There is no rational number $x$ for which $x = n/0$ for some $n$, i.e. $m$ can never be zero.
You also seem to be assuming that values of $m$ and $n$ determine the value of $f$ on the entire interval. This is not true - differing values of $x$ will have differing values of $m$ and $n$, and those $m$ and $n$ values only determine the value of $f$ at that one, single value of $x$.
As to what you say about partitions, I don't really understand it, but again it sounds like you think you can make different choices of $m$ and $n$, and this will lead to different "versions" of the function $f$. This is not the case - $f$ is uniquely defined by what is written.