Comparison
Pointwise convergence means at every point the sequence of functions has its own speed of convergence (that can be very fast at some points and very very very very slow at others).
Imagine how slow that sequence tends to zero at more and more outer points:
$$\frac{1}{n}x^2\to 0$$
Uniform convergence means there is an overall speed of convergence.
In the above example no matter which speed you consider there will be always a point far outside at which your sequence has slower speed of convergence, that is it doesn't converge uniformly.
Another Approach
One can check uniform convergence by considering the "infimum of speeds over all points". If it doesn't vanish then it is uniformly convergent. And that gives another characterization as the ones with nonvanishing overall speed of convergence.
For your third question, JohnD's answer in the page you are quoting, contains a trick that's often used in Analysis courses. That is, try to find:
$$\sup_{x\in[0,1]}|f_n(x)-f(x)|$$
The supmemum occurs at $x$ such that:
$$\frac{df_n(x)}{dx}=0$$
Solving the above for $x$, you get:
$$x=\pm\frac{1}{n}$$
$\frac{1}{n}\in[0,1]$, so substitute back to the function to get:
$$\sup_{x\in[0,1]}|f_n(x)|=f_n\left(\frac{1}{n}\right)=\frac{1}{2}$$
And this is fixed and does not vanish, therefore convergence in $[0,1]$ is not uniform.
Addendum for comment:
I am adding a graphic, so you can see what's happening as a response to your second question.
This is the graph of $f_1(x)$, $f_2(x)$,..., $f_5(x)$, from right to left. Note that the supremum is given by $\left(\frac{1}{n},f_n\left(\frac{1}{n}\right)\right)=\left(\frac{1}{n},\frac{1}{2}\right)$ and is moved to the left on each iteration, but always stays at 1/2.
Best Answer
For pointwise convergence, you fix $x\in [0,1] $ and you compute $\lim_ {n\to+\infty}f(x) $. There will be pointwise convergence in the set containing $x $ for which the limit exists $(\in\mathbb R) $. in your example,
for $x=0$, the limit is zero. for $x=1$, it is zero. for $0 <x <1$, write $(1-x)^n=e^{n\ln (1-x)} $ and you will find zero since exponential is faster than polynomial.
thus all the limits are zero in $[0,1] $ . $(f_n) $ converges (pointwise) to function $0$ at $[0,1] $
For uniform convergence
find the maximum of $|f_n (x)-0|=f_n (x)$ at $[0,1] $.
$$f'_n (x)=n^2 (1-x)^{n-1}(1-x-nx) $$
it is attained at $$x_n=\frac {1}{n+1} $$ $$f (x_n)=n (\frac {n}{n+1})^{n+1}$$
$$\lim_{n\to+\infty}f_n (x_n)=+\infty$$
the convergence is not uniform at $[0,1] $.