Let $A,B,C$ be abelian groups. $A$ has the property that for any homomorphism $\alpha:A\to B$ and surjective homomorphism $\phi:C\to B$ there exists a homomorphism $\beta:A\to C$ such that $\alpha = \phi\circ\beta$. I think this is equivalent to saying that whenever $\phi: C\to B$ is surjective, the induced "push-forward" map on $Hom$ groups $\phi_*:Hom(A,C)\to Hom(A,B)$ is surjective, as $\alpha\in Hom(A,B)$ and $\alpha = \phi_*(\beta)$.
I am trying to show that there exists a free abelian group $F$ and abelian group $D$ such that $F\cong A\oplus D$, and frankly have no idea where to begin. My only thought so far is to try and show that a free resolution of $A$, $0\to F_1 \to F_0 \to A \to 0$ is a split exact sequence, and then $F_0\cong A \oplus F_1$, but I suspect this is not the case as there is no condition on $D$ to be free... any advice would be much apprecaited.
Such objects are called "projective".
Let $F$ be any free group that surjects onto $A$. For example, if $X\subseteq A$ is a generating set for $A$, you can take the free group on $X$. Let $\pi\colon F\to A$ be a surjection.
Now in your condition take $A$ to play the role of $B$, the identity morphism to play the role of $\alpha$, and $\pi$ to play the role of $\phi$. Show that the resulting $\beta\colon A\to F$ is injective, and then check that $D=\ker(\pi)$ will work.