Let $H=\{1,h\}$ and $A=\{0,a\}$ be groups, and $\pi:H\rightarrow \text{Aut}(A)$ be the trivial homomorphism. I have found $FS(H,A,\pi)=\{f_0,f_1\}$ and $IFS(H,A,\pi)={f_0}$
where $f_0(1,1)=f_0(1,h)=f_1(h,1)=f_0(h,h)=0$ and
$f_1(1,1)=f_1(1,h)=f_1(h,1)=0,f_1(h,h)=a$
So I know that we only get two extensions of $H$ by $A$ which is obvious enough, but how can I actually describe the group isomorphic to $Z_4$ by $\frac{FS(H,A,\pi)}{IFS(H,A,\pi)}=EXT(H,A,\pi)=\{f_0,f_1\}$?
Is there a way to find the group presentation for the group by knowing $EXT(H,A,\pi)$?
I am not absolutely certain that the following answer is what you are looking for, but it’s worth a shot.
First, a motivation for the following construction:
If $1 → A \overset ι → G \overset π → H → 1$ is an exact sequence of groups and $σ\colon H → G,~ h ↦ σ_h$ is a section of $π$, i.e. $πσ = \mathrm {id}$, then you can write $$G = ιA·σH \quad \text{with} \quad iA ∩ σH = σ_1,$$ that is: Every element $g ∈ G$ can uniquely be written as $g = a·σ_h$ with some $h ∈ H$ and some $a ∈ ιA = A$ (let’s identify $ιA$ and $A$). Here, $σ$ needs not to be a group homomorphism. (You can think of it as being a choice of representatives for the cosets $G/A \cong H$.)
For $h, h' ∈ H$, notice that $π(σ_h·σ_h') = hh' = π(σ_{hh'})$, so you can uniquely write $$σ_h·σ_{h'} = x_{h,h'} · σ_{hh'}$$ with some $x_{h,h'} ∈ A$. Therefore, if you have elements $g, g' ∈ G$, written as $g = a · σ_h$ and $g' = a'·σ_{h'}$ with $a, a'∈ A$ and $h,h' ∈ H$, then one can carry out the multiplication of $g·g'$ by $$g·g' = (a·σ_h)·(a'·σ_{h'}) = a·(a')_{σ_h}·x_{h,h'} ·σ_{hh'},$$ where $(a')^{σ_h} = σ_h · a' · σ_h^{-1}$ is the conjugation of $a'$ at $σ_h^{-1}$, which is an element in $A$, a normal subgroup of $G$.
Now, $x\colon H × H → A,~〈h,h'〉 ↦ x_{h,h'}$ turns out to be a $2$-cocycle if $A$ is abelian (and if you equip $A$ with the structure of an $H$-module by cojugation $ha := σ_h · a · σ_h^{-1}$). Since the multiplication on $G$ is fully determined by $x$ in terms of multiplication in $A$ and $H$, $x$ is all we need to know about the extension if we know $A$ and $H$.
Starting from a cocycle $f$, the idea is to let it mimic the role of $x$ by an explicit construction.
Now here is the way an extension is constructed from a cocycle:
If $f : H × H → A,~〈h,h'〉 ↦ f_{h,h'}$ is a $2$-cocycle with respect to a linear action of $H$ on $A$, say $H × A → A,~〈h,a〉 ↦ ha$, define on $G = A × H$ a multiplication $G × G → G$ by $$(a,h)·(a',h') = (a · ha' · f_{h,h'},hh').$$ You can use the cocycle relation $hf_{h',h''}·f_{h,h'h''} = f_{hh',h''} · f_{h,h'}$ to see that this is really defining a group where $(f_{1,1}^{-1},1)$ is the neutral element, where the operation of $h$ on $A$ is given – in your case by the trivial morphism $H → \mathrm{Aut}(A)$. The inverse for $(a,h) ∈ G$ is given by $$(h^{-1}a^{-1}·f_{1,1}^{-1}·f_{h^{-1},h}^{-1},h^{-1}).$$
If you take $ι \colon A → G,~a ↦ (a·f_{1,1}^{-1},1)$ and $π\colon G → H,~(a,h) ↦ h$, you indeed get a short exact sequence of groups and group homomorphisms $$1 → A \overset ι → G \overset π → H → 1.$$
Notice that this construction exactly coincides with the construction of the semi-direct product if $f = 1$, constantly. (This happens if and only if the coice of representatives $σ\colon H → G,~h ↦ (1,h)$ is group homomorphism, i.e. if and only if $1 → A → G → H → 1$ is split.)
For $f = f_1$, you can explicitly compute the multiplication table. Because $f_{1,1} = f_1(1,1) = 0$, your neutral element is $(0,1)$ for example.