I don't know from which result this is a corollary in your book, or what you know about modules over a PID, so I will give you a proof from scratch.
Let $M$ be an $R$-module generated by $n $elements, and let $N$ be a submodule of $M.$
Assume that you know the desired result is true when $M$ is free (I will handle this case later). Now, for a general $M$, you have an isomorphism $M\simeq A^n/P$, where $P$ is a submodule of $A^n.$ Then submodules of $M$ correspond via this isomorphism to submodules of $A^n/P$. These submodules have the form $N'/P$ where $N'$ is a submodule of $A^n$ containing $P.$ By the case of free modules, $N'$ is generated by $m\leq n$ elements, and thus so is $N'/P$. Hence, it is also true for the submodules of $M.$
It remains to handle the case where $M$ is free.We will proceed by induction on the rank $n$ of $M$. If $n=0$, then $M=0$ and there is nothing to do. Now assume the result true for any free module of rank $n$, and let $M$ be a free module of rank $n+1$. Fix a basis $(e_1,\ldots,e_{n+1})$ of $M$.
Let $N$ be a submodule of $M$ and let $(x_j)_{i\in J}$ be a family of generators of $N$. We have $$x_{j}=\sum_{i=1}^{n+1} a_{ij}\cdot e_i \mbox{ for all }j\in J.$$
Let $\mathfrak{a}$ be the ideal generated by the $a_{n+1 \ j},j\in J$.Since $R$ is a PID, we have $\mathfrak{a}=(a)$. We may write $$a=\sum_{j\in J} \lambda_j a_{n+1 \ j},$$ where the $\lambda_j's$ are all zero, except for a finite number of them.
Set $$y_0=\sum_{j\in J} \lambda_j\cdot x_j\in N.$$ Then, the $(n+1)$-th coordinate of $y_0$ is $a$.
One may also write $a_{n+1 \ j}=\mu_j a$, and thus the submodule
$N'$ generated by the elementss $x_j-\mu_j y_0,j\in J$ is a submodule of $M'=Re_1\oplus\cdots\oplus Re_n$, which is free of rank $n$. By induction hypothesis, $N'$ is generated by $y_1,\ldots,y_m\in N'\subset N, m\leq n.$
One deduce easily that $x_j$ is a linear combination of $y_0,y_1,\ldots,y_{m}$. Hence, $N$ is generated by the $m+1\leq n+1$ elements $y_0,\ldots,y_{m}$.
This finishes the induction step, and the proof.
Best Answer
A submodule is again a finitely generated module over a PID, hence it is of the form
$N \cong R/q_1^{m_1} \oplus \dotsb \oplus R/q_s^{m_s} \oplus F^\prime$
with $m_i \geq 1$ and $F^\prime$ free.
By tensoring with the quotient field, it is clear that the rank of $F^\prime$ is at most the rank of $F$.
By localizing the injection $N \hookrightarrow M$ at the prime $(q_j)$, we obtain that each $q_j$ occurs within the $p_j$. If not, the right hand side would have killed all torsion, while the left hand side still had torsion.
Thus we can write
$$N \cong R/p_1^{m_1} \oplus \dotsb \oplus R/p_k^{m_k} \oplus F^\prime$$
with $m_i \geq 0$.
By localizing at $(p_i)$ and then passing to the torsion part, we get an injection $R/p_i^{m_i} \hookrightarrow R/p_i^{n_i}$.
In particular $1 \in R/p_i^{m_i}$ is annihilated by $p_i^{n_i}$, hence $p_i^{n_i} \in (p_i^{m_i})$ or equivalently $m_i \leq n_i$.
Summarizing, we get the (intuitively obvious) result, that $N$ is of the form
$$N \cong R/p_1^{m_1} \oplus \dotsb \oplus R/p_k^{m_k} \oplus F^\prime$$
with $0 \leq m_i \leq n_i$ and $\operatorname{rank} F^\prime \leq \operatorname{rank}F$.