Need help with fundamental theorem of homomorphisms 
Theorem (FTH)  Let $f: G \rightarrow H$ be a group homomorphism and let $N$ be a normal subgroup of G which is contained in $\ker(f)$ Then there exists a unique group homomorphism $\overline{f}:G/N \rightarrow H$ such that $\overline{f}\circ v=f$. Here $v: G \rightarrow G/N$ denotes the natural epimorphism. In other words, $\overline{f}(aN)=f(a)$ for all $a \in G$. Moreover, $\operatorname{im}(f)$ and $\ker(\overline{f})= \{ aN \in G/N\mid a\in \ker(f) \}=\ker(f)/N $

Let's start with with the basics. What is $G/N$? and what is a normal subgroup of G which is contained in $\ker(f)$?. I understand what it is, but I feel like I am missing the big picture to really understand this theorem.

On equality $ker(\overline{f})=ker(f)/N$ we see the need to $N$ be contained in $ker(f)$ and, obviously, be normal in $G$. To fix ideas, this result can be understood as follows:

If $f:G\longrightarrow H$ be a group homomorphism and $N$ a normal subgroup of $G$ which is contained in $ker(f)$, then there exists a unique group homomorphism $\overline{f}:G/N\longrightarrow H$ such that the following diagram commutes $$\require{AMScd}\begin{CD}G@>{f}>>H\\@V{v}VV&@VVid_{H}V\\G/N@>>\overline{f}>H\end{CD}$$ where $v$ is the natural epimorphism and $id_{H}$ is the identity of $H$.

In other words, we can decompose $f$ in a unique composition of homomorphisms where one is $v$.
Let's prove it. We want $\overline{f}$ satisfyng $\overline{f}\circ v(x)=\overline{f}(xN)=f(x)$, so define $$\begin{matrix}\overline{f}:&G/N&\longrightarrow&H\\&xN&\longmapsto&f(x)\end{matrix}$$
We need to show that:
$1)$ $\overline{f}$ is well-defined: if $xN=yN$, then $xy^{-1}\in N\subseteq ker(f)$. Thus $$f(x)f(y)^{-1}=f(x)f(y^{-1})=f(xy^{-1})=1_{H}$$ so $f(x)=f(y)$, i.e., $\overline{f}(xN)=\overline{f}(yN)$. So we prove that $\overline{f}$ is well-defined.
$2)$ $\overline{f}$ is a group homomorphism: for all $xN,yN\in G/N$, $$\overline{f}(xN\cdot yN)=\overline{f}(xyN)=f(xy)=f(x)f(y)=\overline{f}(xN)\overline{f}(yN).$$
$3)$ $\overline{f}$ is such that the diagram above commutes: it is follow from the definition.
$4)$ $\overline{f}$ is the unique group homomorphism satisfying this property: if $g:G/N\longrightarrow H$ is such that that diagram above commutes, then for all $xN\in G/N$ we have $$g(xN)=g\circ v(x)=f(x)=id_{H}\circ f(x)=\overline{f}\circ v(x)=\overline{f}(xN).$$ Therefore $g=\overline{f}$.
The proof is done.
