[proofplan] Choose a finite generating set for each submodule $N_i$. The union of these finitely many finite generating sets is finite, so it is a candidate finite generating set for the whole sum. We then take an arbitrary element of $N_1+\cdots+N_k$, write it as a sum of elements from the individual $N_i$, expand each summand using the chosen generators of $N_i$, and collect the resulting left $R$-linear combination. [/proofplan]

[step:Choose finite generating sets for the summands] For each index $i \in \{1, \ldots, k\}$, since $N_i$ is finitely generated as a left $R$-module, choose an integer $n_i \ge 0$ and elements \begin{align*} x_{i,1}, \ldots, x_{i,n_i} \in N_i \end{align*} such that \begin{align*} N_i = Rx_{i,1} + \cdots + Rx_{i,n_i}. \end{align*} If $n_i = 0$, this means $N_i = 0$ and the displayed generating list is empty. Define the finite subset $S \subset M$ by \begin{align*} S := \{x_{i,j} \in M : 1 \le i \le k \text{ and } 1 \le j \le n_i\}. \end{align*} Because the index set of pairs $(i,j)$ with $1 \le i \le k$ and $1 \le j \le n_i$ is finite, the set $S$ is finite. [/step]

[step:Show the combined generators span the finite sum]We prove that $S$ generates $N_1+\cdots+N_k$ as a left $R$-module. Let $y \in N_1+\cdots+N_k$. By the definition of the sum of submodules, there exist elements \begin{align*} y_i \in N_i \end{align*} for $i \in \{1,\ldots,k\}$ such that \begin{align*} y = y_1 + \cdots + y_k. \end{align*} For each $i$, since $x_{i,1}, \ldots, x_{i,n_i}$ generate $N_i$, there exist coefficients \begin{align*} r_{i,1}, \ldots, r_{i,n_i} \in R \end{align*} such that \begin{align*} y_i = r_{i,1}x_{i,1} + \cdots + r_{i,n_i}x_{i,n_i}. \end{align*} Substituting these expressions into the decomposition of $y$ gives \begin{align*} y = \sum_{i=1}^{k}\sum_{j=1}^{n_i} r_{i,j}x_{i,j}. \end{align*} Thus $y$ is a left $R$-linear combination of elements of $S$. Since $y$ was arbitrary, $S$ generates $N_1+\cdots+N_k$.[/step]

[guided]The goal is to prove finite generation of the sum, so we must exhibit one finite list of elements that spans it. The natural candidate is the list obtained by putting together all the generators of the individual submodules. Let $y \in N_1+\cdots+N_k$. The definition of the submodule sum says exactly that $y$ can be decomposed as \begin{align*} y = y_1 + \cdots + y_k, \end{align*} where each summand satisfies $y_i \in N_i$. Now fix an index $i \in \{1,\ldots,k\}$. Since $x_{i,1}, \ldots, x_{i,n_i}$ generate $N_i$ as a left $R$-module, the element $y_i \in N_i$ can be written with coefficients on the left: \begin{align*} y_i = r_{i,1}x_{i,1} + \cdots + r_{i,n_i}x_{i,n_i} \end{align*} for some $r_{i,1}, \ldots, r_{i,n_i} \in R$. The coefficients are placed on the left because $M$ is a left $R$-module. Substituting these expressions for all $i$ into the decomposition of $y$ gives \begin{align*} y = \sum_{i=1}^{k}\sum_{j=1}^{n_i} r_{i,j}x_{i,j}. \end{align*} Every element $x_{i,j}$ appearing in this expression belongs to the set \begin{align*} S = \{x_{i,j} \in M : 1 \le i \le k \text{ and } 1 \le j \le n_i\}. \end{align*} Therefore $y$ is a left $R$-linear combination of elements of $S$. Since the argument applies to every $y \in N_1+\cdots+N_k$, the set $S$ spans the whole sum.[/guided]

[step:Conclude finite generation of the sum] The set $S$ is finite and generates $N_1+\cdots+N_k$ as a left $R$-module. Therefore $N_1+\cdots+N_k$ is finitely generated as a left $R$-module. [/step]