Let $p$ be a prime. Let $f: \mathbb{Z}_p \to \mathbb{Z}_p$ be surjective modulo $p^n, \forall n \in \mathbb{N}$, where $\mathbb{Z}_p$ is the set of $p$-adic integers.
If it is given that $g: \{0,1,\ldots,p-1\} \to \{0,1,\ldots,p-1\}$ is defined as
$$g(i) = \frac{f(m+ ip^{1+\lfloor \log_p m\rfloor }) - f(m)}{p^{1+ \lfloor \log_p m\rfloor }} \text{ (mod } p) $$
How do I show that $g$ is also surjective?
I am planning to show $g$ represents all elements of $\{0,1,\ldots,p-1\}$ so that it is onto.
However $g$ being a modulo of a fraction makes it confusing for me.
Without further conditions on $f$, and the way you want the dependence on $m$ be interpreted as per comment, this is wrong. Example: Let $p \neq 2$ and
$$f(x) := \begin{cases} \dfrac{1}{p}\log(x) \quad \text{ if } x \equiv 1 \text{ mod } p \\ 0 \qquad \qquad \text{ otherwise } \end{cases}$$
Then if $m \not\equiv 1$ mod $p$, the formula gives $g(i)=0$ for all $i$.
Some things are strange in the question , namely, $f$ being "surjective mod $p^n$ for all $n$" is the same as just saying it is surjective; and I still find the dependence on $m$ weird, as noted in a comment.
Also, the formula for $g$ looks like a difference quotient, but I do not see how that would be of any help, why would anyone be interested in surjectivity mod $p$ of that expression? Maybe can you give some context and motivation?