Constructing a group action from a group extension + a section Let $F,G$ be two groups. An extension of $G$ by $F$ is a triple $\mathscr{E}=(E,i,p)$, where $E$ is a group, $i:F\rightarrow E$ is an injective homomorphism, and $p:E\rightarrow G$ is a surjective homomorphism such that $Im(i)=Ker(p)$. 
Denote by $\mathscr{E}:F\xrightarrow{i} E\xrightarrow{p} G$ the extension $\mathscr{E}=(E,i,p)$ of $G$ by $F$.
Let $\mathscr{E'}:F\xrightarrow{i'} E'\xrightarrow{p'} G$ be an extension of $G$ by $F$ and $s':G\rightarrow E'$ a section of $\mathscr{E'}$ (i.e., $s'$ is a homomorphism such that $p'\circ s'=id_G$).
Claim. Let $Int(s'(g))$ denote the inner automorphism defined by $s'(g)$. The following formula defines an action $\tau$ of $G$ on $F$:
$$i'(\tau(g,f))=s'(g)i'(f)s'(g)^{-1}=Int(s'(g))(i'(f)).$$
(Bourbaki Algebra Chapter 1, § 6, no. 1)
Question: Is this a legitimate way of defining an action of $G$ on $F$? I mean can one define a mapping simply by specifying how it behaves when composed with another mapping? This doesn't seem to tell me how $G$ acts on elements of $F$...
 A: The short answer is yes -- this does define a group action of $G$ on $F$. I think the big idea that you are currently missing is that $i$ is actually an isomorphism, so that we can transfer the action defined on the image of $i$ in $E'$ back onto $F$.
I am going to use the slightly more suggestive names $K$ for kernel and $Q$ for quotient throughout this answer. I will also use some terminology that is standard when working with group extensions, but I'll define everything as we go.

Say we have a group extension as follows (if you aren't familiar with exact sequences, this just says that $i$ is injective and $p$ is surjective):
$$0 \to K \xrightarrow{i} G \xrightarrow{p} Q \to 0$$
This tells us that we can pretend $K \leq G$, since $K \cong i(K) \leq G$. So we we are justified in saying $G/K \cong Q$.
Moreover, let's say that this exact sequence is split -- by this I mean there is a group homomorphism $s : Q \to G$ such that $ps = \text{id}_Q$. How can we define a group action of $Q$ on $K$?
For any $g \in G$, let's define
$$\gamma_g(k) = gkg^{-1}.$$
Notice we have $gkg^{-1} \in K$ since $K \trianglelefteq G$, so these are all automorphisms of $K$. It is quick to check that $\gamma_g \gamma_h k = \gamma_{gh} k$, 
so this is an action of $G$ on $K$.
Of course we want an action of $Q$ on $K$, so what should we do? Well we know that 
$s : Q \to G$ is a homomorphism, so let's define an action by taking 
(for every $q \in Q$):
$$\gamma_{s(q)} : K \to K$$
In the notation of an action, this says 
$$q \cdot k = s(q) k s(q)^{-1}$$

But, I hear you asking, why is this allowed? Have we not really defined an action on $i(K) \leq G$? How do we know that this is an action on $K$? The reason is that $i$ is an isomorphism between $K$ and $i(K)$. It is injective by assumption, and it is surjective because we restrict to $i(K)$ instead of $G$. So if you want to be extra pedantic, we could argue as follows:
We have an action $\gamma_{s(q)} : i(K) \to i(K)$. So we define $\tau_q : K \to K$ by 
$$\tau_q(k) = i^{-1}(\gamma_{s(q)}(i(k)))$$
Here $i^{-1}$ is well defined since $i$ is an isomorphism, and it is fairly quick to see that this is still a group action. Expanding out the above definition of $\tau$ will give the same one from your question, so you can check that we haven't cheated.

I hope this helps ^_^
