I'm using the lifting definition of projective for the purposes of this question. That is a module $M$ is projective if
$$\begin{matrix} & & M & \\ & & \downarrow &\\ B &\rightarrow & C &\rightarrow &0 \end{matrix}$$
for any surjective $g: B\rightarrow C$ and $h: M\rightarrow C$, there exists a $p : M\rightarrow B$ such that $gp = h$.
In the proof that a free module is projective, we take a basis $\{m_i\}$ for $M$. Then since $g$ is surjective, there exists $b_i$ such that $g(b_i) = h(m_i)$. Then we define $p(m_i) = b_i$ and this extends to the homomorphism that we want.
My question is why this argument doesn't work for an arbitrary module $M$ by taking a generating set instead of a basis? I know it doesn't work but I can't see what goes wrong in the argument.
Because there might be a relations between the elements of the generating set that may not be preserved by the $p$ as defined. For example if $\{m_i\}$ is a generating set for $M$ and $r_i \in R$ (finitely many non-zero) is such that $\Sigma r_i m_i = 0$. Then we would expect $p(\Sigma r_i m_i)=\Sigma r_i p(m_i) = \Sigma r_i b_i$ to be zero which may not necessarily hold in $B$. Jyrki posted a nice counterexample in the comments