[proofplan]
We prove that the coordinate map determined by the ordered basis is a bijective homomorphism of left $R$-modules. The homomorphism property follows from the coordinatewise module structure on $R^n$ and the distributive and associative laws in the left $R$-module $M$. Surjectivity is exactly the spanning part of the basis property, while injectivity is exactly the linear independence part.
[/proofplan]
[step:Verify that the coordinate formula defines a left $R$-module homomorphism]
For each element $(r_1,\ldots,r_n) \in R^n$, the expression $\sum_{i=1}^{n} r_i b_i$ is a finite sum in the left $R$-module $M$, so $\varphi$ is a well-defined map from $R^n$ to $M$.
Let $x = (r_1,\ldots,r_n)$ and $y = (s_1,\ldots,s_n)$ be elements of $R^n$. Since addition in $R^n$ is coordinatewise, $x+y = (r_1+s_1,\ldots,r_n+s_n)$. Using distributivity in the left $R$-module $M$, we obtain
\begin{align*}
\varphi(x+y) = \sum_{i=1}^{n} (r_i+s_i)b_i.
\end{align*}
For each $i$, distributivity gives $(r_i+s_i)b_i = r_i b_i + s_i b_i$, so
\begin{align*}
\varphi(x+y) = \sum_{i=1}^{n} r_i b_i + \sum_{i=1}^{n} s_i b_i.
\end{align*}
By the definitions of $\varphi(x)$ and $\varphi(y)$, this is $\varphi(x)+\varphi(y)$. Let $a \in R$. Since scalar multiplication on the left $R$-module $R^n$ is coordinatewise, $a x = (a r_1,\ldots,a r_n)$. Using associativity of scalar multiplication in $M$, we obtain
\begin{align*}
\varphi(a x) = \sum_{i=1}^{n} (a r_i)b_i.
\end{align*}
For each $i$, associativity gives $(a r_i)b_i = a(r_i b_i)$, so
\begin{align*}
\varphi(a x) = \sum_{i=1}^{n} a(r_i b_i).
\end{align*}
By distributivity over finite sums in $M$, this is
\begin{align*}
\varphi(a x) = a\sum_{i=1}^{n} r_i b_i.
\end{align*}
Using the definition of $\varphi(x)$, this is $a\varphi(x)$. Thus $\varphi$ preserves addition and left scalar multiplication, so $\varphi$ is a homomorphism of left $R$-modules.
[guided]
We first check the algebraic structure before using the basis property. The domain $R^n$ is the free left $R$-module with coordinatewise operations: for $x = (r_1,\ldots,r_n)$ and $y = (s_1,\ldots,s_n)$ in $R^n$, their sum is
\begin{align*}
x+y = (r_1+s_1,\ldots,r_n+s_n),
\end{align*}
and for $a \in R$, scalar multiplication is
\begin{align*}
a x = (a r_1,\ldots,a r_n).
\end{align*}
The codomain $M$ is a left $R$-module, so each product $r_i b_i$ lies in $M$, and the finite sum $\sum_{i=1}^{n} r_i b_i$ is an element of $M$. Hence the displayed formula defines a map $\varphi: R^n \to M$.
To prove additivity, take arbitrary elements $x = (r_1,\ldots,r_n)$ and $y = (s_1,\ldots,s_n)$ of $R^n$. Applying the definition of $\varphi$ to the coordinatewise sum gives
\begin{align*}
\varphi(x+y) = \sum_{i=1}^{n} (r_i+s_i)b_i.
\end{align*}
For each index $i$, distributivity of the module action over addition in $R$ gives $(r_i+s_i)b_i = r_i b_i + s_i b_i$. Summing these equalities in $M$ gives
\begin{align*}
\varphi(x+y) = \sum_{i=1}^{n} r_i b_i + \sum_{i=1}^{n} s_i b_i.
\end{align*}
By the definition of $\varphi(x)$ and $\varphi(y)$, this is exactly $\varphi(x)+\varphi(y)$.
To prove compatibility with left scalar multiplication, take an arbitrary scalar $a \in R$ and an arbitrary element $x = (r_1,\ldots,r_n) \in R^n$. The scalar multiple in $R^n$ is $a x = (a r_1,\ldots,a r_n)$, so
\begin{align*}
\varphi(a x) = \sum_{i=1}^{n} (a r_i)b_i.
\end{align*}
Because $M$ is a left $R$-module, the associativity axiom for scalar multiplication gives $(a r_i)b_i = a(r_i b_i)$ for each $i$. Therefore
\begin{align*}
\varphi(a x) = \sum_{i=1}^{n} a(r_i b_i).
\end{align*}
Distributivity of scalar multiplication over finite sums in $M$ gives
\begin{align*}
\varphi(a x) = a\sum_{i=1}^{n} r_i b_i.
\end{align*}
Using the definition of $\varphi(x)$, this becomes
\begin{align*}
\varphi(a x) = a\varphi(x).
\end{align*}
Thus $\varphi$ preserves both addition and left scalar multiplication, which is precisely the condition that $\varphi$ be a homomorphism of left $R$-modules.
[/guided]
[/step]
[step:Use the spanning property of the basis to prove surjectivity]
Let $m \in M$. Since $B = \{b_1,\ldots,b_n\}$ is a module basis of $M$, it spans $M$ over $R$. Hence there exist coefficients $r_1,\ldots,r_n \in R$ such that
\begin{align*}
m = \sum_{i=1}^{n} r_i b_i.
\end{align*}
For $x = (r_1,\ldots,r_n) \in R^n$, the definition of $\varphi$ gives $\varphi(x) = m$. Since $m \in M$ was arbitrary, $\varphi$ is surjective.
[/step]
[step:Use linear independence of the basis to prove injectivity]
Let $x = (r_1,\ldots,r_n) \in R^n$ satisfy $\varphi(x)=0_M$, where $0_M$ denotes the zero element of $M$. By the definition of $\varphi$,
\begin{align*}
\sum_{i=1}^{n} r_i b_i = 0_M.
\end{align*}
Since $B$ is a module basis, it is linearly independent over $R$. Therefore each coefficient in this finite linear relation is zero:
\begin{align*}
r_i = 0_R \quad \text{for every } i \in \{1,\ldots,n\},
\end{align*}
where $0_R$ denotes the zero element of $R$. Hence $x = (0_R,\ldots,0_R)$, the zero element of $R^n$. Therefore the only element of $R^n$ mapped by $\varphi$ to $0_M$ is the zero element of $R^n$, so $\varphi$ is injective.
[/step]
[step:Conclude that the coordinate map is an isomorphism]
We have shown that $\varphi: R^n \to M$ is a homomorphism of left $R$-modules, that $\varphi$ is surjective, and that $\varphi$ is injective. Therefore $\varphi$ is a bijective left $R$-module homomorphism. By the definition of an isomorphism of left $R$-modules, $\varphi$ is an isomorphism.
[/step]
