At the very beginning of Serre's Trees, it's taken that the groups $G_i$ (indexed over some set $I$, with no additional specifications) are equipped with homomorphisms $f_{ij}:G_i\to G_j$, collected in $F_{ij}$ defined for all $i,j\in I$. From this, he constructs the direct limit $G$ with maps $f_i:G_i\to G$, satisfying $f_j\circ f_{ij}=f_i$ and a universal property. Then, he gives the example $I=[2]$, with the only specifications on the groups $G_1$ and $G_2$ being the existence of homomorphisms $f_i:A\to G_i$ for some other group $A$, and claims that $G$ in this setting is the amalgamated product $G_1*_AG_2$.
My question is exactly what implicit information about $F_{12}$ and $F_{21}$ is given in that example. Here is my best guess. Say $G_1$ is generated by (fixed) $S$, and $T\subset A$ has the set bijection $\phi:T\overset{\sim}{\to}S$ where $\phi=f_1|_T$. Then $f_T=f_2\circ\phi^{-1}$, which extends to all of $G_1$, and $F_{12}=\{f_T:T\subset A,f_1|_T:T\overset{\sim}{\to}S\}$. $F_{21}$ arises analogously. I think this is equivalent to the notion of amalgamation that I've seen elsewhere of handling $f_1(x)f_2(x)^{-1}$, but wanted to be certain by phrasing it explicitly in the notation Serre introduces for the direct limit. Thanks in advance.
Okay, just looked at the book. So for each pair of indices $i$ and $j$; we have a family $F_{ij}$ of morphisms. Far be it for me to correct Serre, so let me instead point out that most people would not call this construction a "direct limit", but rather a colimit. "Direct limit" is usually restricted for the case where $I$ is a directed set (partially ordered set in which every finite subset has an upper bound), and where for each $i$ and $j$, if $i\nleq j$ then $F_{ij}$ is empty; if $i\leq j$ then $F_{ij}=\{f_{ij}\}$ is a singleton; $f_{ii}=\mathrm{Id}_{G_i}$; and the morphisms are requied to satisfy $f_{ik} = f_{jk}\circ f_{ij}$ whenever $i\leq j\leq k$.
I don't think you are reporting the example correctly: the index set is not "$[2]$" as you report (by which I suspect you mean $\{1,2\}$). Here is what he says, explicitly:
So, here your index set has three elements: $1$, $2$, and then a silent index used for $A$. I would denote it by $G_0=A$. Then you have $F_{01}=\{f_1\colon A\to G_1\}$; $F_{02}=\{f_2\colon A\to G_2\}$; and $F_{10}=F_{20}=F_{12}=F_{21}=\varnothing$. In other words, you entire system consists of exactly three groups and exactly two maps, and nothing else. Then you take the colimit/direct limit of this system, and you get the free product of $G_1$ and $G_2$ amalgamated over $A$.
Note that $f_1$ and $f_2$ are not assumed to be embeddings; that's why the amalgamated product could be trivial.
Your $S$ and $T$ are misguided. If $G_1$ were generated by $S$ and you had a bijection from a subset of $A$ onto $S$ with the restriction of a homomorphism, then the homomorphism would be onto.
The "classic" situation is when $f_1$ and $f_2$ are embeddings, so that $A$ can be identified with a "common" subgroup of $G_1$ and $G_2$; in that case, $G_1*_A G_2$ is defined as $G_1*G_2/N$, where $G_1*G_2$ is the free product, and $N$ is the smallest normal subgroup of $G_1*G_2$ containing all elements of the form $f_1(a)f_2(a)^{-1}$; it is then necessary to prove that this results in a group that contains isomorphic copies of $G_1$ and $G_2$, which intersect precisely "at" $A$. In the more general situation here, where $f_1$ and $f_2$ are not required to be monomorphisms/injective, this is what is usually called a pushout of $f_1$ and $f_2$; that is, you have a diagram $$\require{AMScd} \begin{CD} A@>f_1>>G_1\\ @Vf_2VV\\ G_2 \end{CD}$$ and then you seek a universal group $G$ and maps $g_1,g_2$ that fits into the diagram as follows: $$\begin{CD} A@>f_1>>G_1\\ @Vf_2VV@VVg_1V\\ G_2@>g_2>>G \end{CD}$$ so that $g_1f_1=g_2f_2$ and satisfies the relevant universal property relative to that condition.