Prove that $Hom_{R}(M,N)\cong M^*\otimes_{R} N$ as groups So the statement of the problem is thus:

Let $R$ be a ring with identity such that $1_{R} \ne 0_{R}$, $M$ be a free left $R$-module of finite rank, and $N$ be a left $R$-module. Then $M^*=Hom_{R}(M,R)$ is a right $R$-module under $(fr)(m)=f(m)r$ for $f\in M^*$, $r\in R$, and $m\in M$. Prove that $Hom_{R}(M,N)$ is isomorphic to $M^*\otimes_{R}N$ as groups.

My sketch of the proof so far is as follows:
Let $\lbrace{m_1,m_2,m_3,...,m_l\rbrace}$ be a basis of $M$. Then $M^*$ has a dual basis given by $\lbrace{f_1,f_2,f_3,...,f_l\rbrace}$. We also know that $M^*\otimes_{R}N$ has elements of the form $\Sigma_i f_i\otimes n_i$. Since $f\in M^*$ as defined by $(fr)(m)=f(m)r$ it follows that each element of $M^*\otimes_{R}N$ can be rewritten as
$$\Sigma_{i=1}^{l} (f_i r)(m_i)\otimes n_i=\Sigma_{i=1}^{l} f_i(m_i)r\otimes n_i = (\Sigma_{i=1}^{l} f_i(m_i)\otimes n_i)r$$
From here I am not entirely sure what to do. Any suggestions would be greatly appreciated!

Your ideas are basically fine.
The dual basis $(f_i)\in M^*$ of the basis $(m_i)\in M$ satisfies $f_i(m_j)=\delta_{i,j}$.
We need to establish an isomorphism between the Abelian groups $\hom_R(M,N)$ and $M^*\otimes_R N$.
The observation that each element of the latter can be written as $\sum_i f_i\otimes n_i$ is a good starting point.
What $M\to N$ $R$-homomorphism should we assign to it?
Well, the map $m\mapsto \sum_i f_i(m)\cdot n_i$. 
Verify that it's a homomorphism of left modules $M\to N$.
For the converse, given $\varphi:M\to N$, consider the element $\sum_i f_i\otimes \varphi(m_i)$ of $M^*\otimes N$.
