Your reasoning that your sequence $(f_n)$ converges pointwise to the zero function $f(x)=0$ on $[0,2]$ is correct.
Considering uniform convergence now, let's first
recall the definition of uniform convergence of a sequence of functions:
The sequence $(g_n)$ converges uniformly to $g$ on $I$ if for every
$\epsilon>0$ there is a positive integer $N$ such that $$\tag{1}
|g_n(x)-g(x)|<\epsilon,\quad\text{for all } n\ge N\ \text{and}\ x\in I. $$
Note that $(1)$ must hold for each $n\ge N$ and every $x\in I$. (It does have something to do with the supremum $\sup\limits_{x\in I}|g_n(x)-g(x)|$: this must be be able to be made as small as desired by taking $n$ sufficiently large.)
Back to your sequence. Note that
for any positive integer $n$, we have $$\tag{2}f_n(1/n)=n.$$
Now let $\epsilon=1 $ and let $N$ be any fixed positive integer. By $(2)$, we have $$|f_N(1/N) -f(1/N)|=f_N(1/N)=N\ge {1 }=\epsilon.$$
This shows that for $\epsilon=1$ there is no positive integer $N$ such that $(1)$ holds (with $g_n=f_n$ and $g=f$). Consequently, $(f_n)$ does not converge uniformly on $[0,2]$ (or in fact, on any interval containing $0$).
The graphs of the $f_n$ prove illuminating.
The sequence resembles the "witch hat" sequence: it consists of spikes near $x=0$ whose widths approach zero but whose heights grow large. The graphs of the first few terms of the sequence are shown below:
