Let $Q$ a left $R$-module, we have that are equivalent:
a. If $X$ and $Y$ are left $R$-modules and $f : X \rightarrow Y$ is an injective module homomorphism and $g : X \rightarrow Q$ is an arbitrary module homomorphism, then there exists a module homomorphism $h : Y \rightarrow Q$ such that $hf = g$
b. Given $0\rightarrow X \xrightarrow{f} Y \xrightarrow{g} Z \rightarrow 0$ an exact short sequence, then $$0\rightarrow Hom_R(Z,Q) \xrightarrow{g^*} Hom_R(y,Q) \xrightarrow{f^*} Hom_R(X,Q) \rightarrow 0$$ is exact.
I've proven that $b \Rightarrow a$ but on the other direction I don't know how to prove that $Ker f^* \subseteq Im g^*.$ I don't know how to approach this.
edit: I made a mistake.
To prove that $Ker f^*\subset Im g^*$ you don't need condition a. Let $h:Y\to Q$ be in the kernel of $f^*$, that is $h\circ f=0$. Then by the universal property of the cokernel of $f$, there is a unique $k:Z\to Q$ such that $k\circ g=h$, i.e. $h$ is in the image of $g^*$.
More generally, the last sequence you wrote is always exact at the first two Hom spaces. Condition a is only telling you that $f^*$ is surjective.