Let $R$ be a reduced Noetherian ring and $M$ be a finitely generated $R$ module of rank $n.$ Let $\operatorname{Ass}(M)\subset \operatorname{Min}(R).$
How to show that
there exists an injective $R$-linear map $f:M\longrightarrow R^n$
I think if the existence of a map from $M$ to $R^n$ can be shown then injectivity comes from the fact that $R$ is reduced and $\operatorname{Ass}(M)\subset \operatorname{Min}(R).$
Let $S$ be the set of non-zero divisors of $R$. Note the assumption implies that $M \to S^{-1}M$ is injective, because $S = R \setminus \bigcup_{P \in Min(R)} P$ does not meet any associated primes of $M$.
By assumption $S^{-1}M$ has a $S^{-1}R$-basis $e_1, \dotsc, e_n$ and w.l.o.g we have $e_i \in M$.
Now consider the map $f:M \to S^{-1}M$ and let $m_1, \dotsc, m_t \in M$ be generators of $M$. We have $m_j = \sum\limits_{i=1}^n \frac{a_{ij}}{s_j}e_i$ for and $f(M) \subset \sum\limits_{i=1}^n R\frac{e_i}{s}$ for $s = s_1 \dotsb s_t \in S$. But this sum is clearly direct, since the $e_i$ are $S^{-1}R$-linear independent.
Thus we have shown that $f$ is an injective map $M \to \oplus_{i=1}^n R\frac{e_1}{s} \cong R^n$.
One can rephrase this for the more experienced user: By assumption we find some non-zerodivisor $t \in R$ an isomorphism $M_t \xrightarrow{\cong} (R_t)^n$, which yields injective map $M \to M_t \to (R_t)^n$. By the finite generation of $M$, we deduce that the image is actually contained in $(t^{-m}R)^n$ for some $m \geq 0$. Hence we have an injection $M \to M_t \to (t^{-m}R)^n \cong R^n$.