What is the main difference between pointwise and uniform convergence as defined here?

$f_n$ converges pointwise means for every $c>0$ for every $x$, there exists $N(x)$ such that $n>N(x)$ implies that $|f_n(x)-f(x)|<c$

$f_n$ converges uniformly means that for every $c>0$ there exists $N$ such that for every $x$, $n>N$ implies that $|f_n(x)-f(x)|<c$.

In the simply convergence, $N(x)$ depends of $x$ but for uniformly convergence one $N$ is chosen for every $x$.

Uniform convergence is actually $\mathcal L^\infty$ convergence, i.e. $$ f_n \rightrightarrows f [x \in E]\!\! \iff \!\! \sup_{x \in E} \vert f_n - f\vert(x) \to 0[n \to \infty]. $$ This is strictly stronger than pointwise convergence.

Alternatively, uniform convergence implies pointwise convergence, so $f_n \to f$ in both cases.

Sorry, but yes, you probably are missing something important, because the second statement in your post

On the other hand $\{f_n\}_{n\in\mathbb{N}}$, converges uniformly to $f$ on $E$ if and only if $$f_n(x)\to f(x),\;\forall\,x\in E.$$

is false. Without seeing your source, it's impossible to say what happened here, where this erroneous statement came from, and what exactly you're missing.

Are you sure the source says if and only if here? This certainly is NOT the definition of uniform convergence (unlike your first statement, which indeed is a definition of pointwise convergence, unless we want to expand it further in $\varepsilon/\delta$-language). This could be a theorem that states that uniform convergence implies pointwise convergence, which is a true theorem, but ONLY in this direction, so it cannot say if and only if.

Your best bet is to check the original source to find out what exactly it says there.