Let $R$ be a ring, $F$ a nonzero free $R$-module, and let $\varphi : M \to N$ be a homomorphism of $R$-modules. Prove that $\varphi$ is onto if and only if for all $R$-module homomorphisms $\alpha : F \to N$ there exists an $R$-module homomorphism $\beta : F \to M$ such that $\alpha=\varphi \circ \beta$.
To show $\varphi$ is surjective, we let $F=R^{\oplus N}$ and consider the canonical projection $\pi: R^{\oplus N} \to N$. For $\pi$ there exists $\beta: R^{\oplus N} \to M$ such that $\pi = \varphi \circ \beta$. Since the natural inclusion $i: N \to R^{\oplus N}$ is an $R$-module hom, then $\mathrm{Id}_N = \pi \circ i= \varphi \circ \beta \circ i$. Thus, $\varphi$ has a right inverse and so it is surjective.
I am stuck as far as the other direction. Any help?
As noted in comments $F$ is fixed, you cannot just choose it as you want.
Assume that $A\subseteq F$ is a (nonempty) free basis.
"$\Leftarrow$" Assume that $\varphi$ is not onto. Let $x\in N$ be such that $x\not\in im(\varphi)$. Note that this also means that $x\not\in im(\varphi\circ\beta)$ for any $\beta$. Now take $\alpha:F\to N$ by definining $\alpha(a)=x$ for some (all) $a\in A$. Since $A$ is a basis then $\alpha$ is a well defined and unique homomorphism. But then $x\in im(\alpha)$ even though $x\not\in im(\varphi\circ\beta)$ and so there's no chance for $\alpha=\varphi\circ\beta$ to hold, regardless of what $\beta$ is.
This contradiction means that $\varphi$ has to be onto. $\Box$
"$\Rightarrow$" So $\varphi:M\to N$ is a given onto homomorphism and $\alpha:F\to N$ is an arbitrary homomorphism. Since $\varphi$ is onto then for any $a\in A$ there is $m_a\in M$ such that $\varphi(m_a)=\alpha(a)$. Now let $\beta:F\to M$ be given by $\beta(a)=m_a$ for $a\in A$. Since $A$ is a free basis then $\beta$ is well defined and unique.
Now for any $a\in A$ we have:
$$(\varphi\circ\beta)(a)=\varphi(\beta(a))=\varphi(m_a)=\alpha(a)$$
and so $\varphi\circ\beta$ and $\alpha$ are equal on $A$. Since $A$ is a basis then they are equal on whole $F$. $\Box$