Let $\phi: G \to H$ be a group homomorphism with $K = \ker\phi$. Then $G/K \simeq \phi[G]. $
The hinge to the proof is to define $\Phi: G/K \to \phi[G]$ given by $\Phi(gK) = \phi(g)$. Then we must prove $\Phi$ an isomorphism and well-defined. I can do this, hence not asking about proofs or formality.
(1.) What’s the intuition? This has a picture but I'm still confounded.
(2.) Where did $\Phi(gK) = \phi(g)$ spring from? I want to understand this, not memorize it.
(3.) How is $\Phi(G/K) = \{ \Phi(gK) : g \in G \}$?
I know definitions $G/K = \{gK : g \in G \}$. For any function $f$, $f[S] = \{ f(s) : s \in S \}$
The best way to see the problem is too show that there is one unique homomorphism $\Phi :G/K\rightarrow \phi[G]$ such that $\Phi\circ\nu=\phi$.
If $\Phi$ is a solution, then you have $\phi(g)=\Phi(\nu(g))=\Phi(gK)$.
You must also show that it defines a function. If $gK=g'.K$, then $\phi(g.g'^{-1})=e'$ and $\phi(g)=\phi(g')$. So the definition of $\Phi(gK)$ do not change if you take $g'$ such as $g'K=gK$.
$\Phi$ is obviously surjective, because $\phi$ is.
$\Phi$ is injective : if $\Phi(gK)=\Phi(hK)$ then $\phi(g)=\phi(h)$ and $g.h^{-1}\in K$, so $gK=hK$.
So $\Phi$ is an isomorphism between $G/K$ and $\phi[G]$
So the formula of the $(2)$ is the only one to be possible.
For the $(3)$, $\phi^{-1}[\{\phi(g)\}]= \{h\in G, \phi(h)=\phi(g)\}=\{h\in G, \phi(h.g^{-1})=e'\}$.
So it is :$\{h\in G, h.g^{-1}\in K\}$ id est $g.K$.
For the $(4)$, with your definitions, $\Phi(G/K)=\{\Phi(x), x\in G/K\}$. But $\{x\in G/K \}=\{g.K, g\in G\}$, so $\Phi(G/K)=\{\Phi(g.K), g\in G\}$.