Why $f\colon \mathbb{Z}_n^\times \to \mathbb{Z}_m^\times$ is surjective?

When $n$ is a power of a prime, it's easy because $$ \gcd(a,p^k)>1 \implies \gcd(a,p)>1 $$

In the general case, use the Chinese Remainder Theorem.

Let $n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ and $m=p_1^{\beta_1}\cdots p_k^{\beta_k}$, with $\alpha_i\geq \beta _i$. Then $$\Bbb Z_n^\times =\prod^{k}_{i=1} \Bbb Z_{p_i^{\alpha_i}}^\times\text{ and }\Bbb Z_m^\times = \prod^{k}_{i=1} \Bbb Z_{p_i^{\beta_i}}^\times$$

The surjectivity of $\Bbb Z_m^\times \to \Bbb Z_n^\times$ follows from the surjectivity of $\Bbb Z^\times_{p_i^{\alpha_i}}\to\Bbb Z^\times_{p_i^{\beta_i}}$.

I don't know an elementary answer, but it is true that any ring surjection $f : R \to S$ where $R$ is a finite ring, induces a surjective group homomorphism $f^\times : R^\times \to S^\times$. For a proof, see my answer on MO (which has a very short proof of a more general statement, using only the Chinese Remainder Theorem).