Let $G$ and $H$ be groups. Their free product is a group $G*H$ with homomorphisms $\iota_G:G\rightarrow G*H$ and $\iota_H:H\rightarrow G*H$ so that given any other group $X$ with homomorphisms $f_G:G\rightarrow X$ and $f_H:H\rightarrow X$ there is a unique homomorphism $f_G*f_H:G*H\rightarrow X$.
Does this mean that as long as we have maps $G\to X$ and $H \to X$ then any non trivial map from $G*H\to X$ has to be the unique map we have above? There can be no other non trivial map $G*H\to X$? Or is it that there is a unique map that makes $f_G*f_H\circ \iota_H =f_H$ and $f_G*f_H\circ \iota_G=f_G$ but other maps that don't make these diagrams commute can exists between $G *H$ and $X$?
You have left out an important part of the definition! The defining property of $G*H$ (together with the homomorphisms $\iota_G$ and $\iota_H$) is that given any group $X$ and homomorphisms $f_G:G\to X$ and $f_H:H\to X$, there is a unique homomorphism $f:G*H\to X$ such that $f\circ\iota_G=f_G$ and $f\circ\iota_H=f_H$. Here "there exists a unique $f$ such that ..." means that there is exactly one ($f$ with this property), not that there exists exactly one $f$ and that one $f$ happens to have this property.
So there will typically actually be many different homomorphisms $G*H\to X$: there is one homomorphism for each pair $(f_G,f_H)$. The homomorphism $f$ is only unique among homomorphisms with the special property that $f\circ\iota_G=f_G$ and $f\circ\iota_H=f_H$.
Note moreover that if you start with any $f$, there do always exist an $f_G$ and $f_H$ such that (in your notation) $f=f_G*f_H$: just define $f_G=f\circ\iota_G$ and $f_H=f\circ\iota_H$. So there really is a bijection between the set of homomorphisms $G*H\to X$ and the set of pairs of homomorphisms $G\to X$ and $H\to X$, given by sending $f$ to $(f_G,f_H)$.