For $x < 2$, since $y=2^{x}$, we have $0 < 2^{x} < 4$.
- When $0 < a < 4$, the equation $|2^{x}-a|=0$ has one solution.
- When $a \leqslant 0$ or $a \geqslant 4$, the equation $|2^{x}-a|=0$ has no solution.

For $x\geqslant 2$, notice that $x^{2}-3ax+2a^{2}=(x-a)(x-2a)$.
- When $a \in (0,1)$, the equation $x^{2}-3ax+2a^{2}=0$ has no solutions in $[1,+\infty)$.
- When $a \in [1,2)$, the equation $x^{2}-3ax+2a^{2}=0$ has exactly one solution in $[1,+\infty)$.
- When $a \in [2,+\infty)$, the equation $x^{2}-3ax+2a^{2}=0$ has exactly two solutions in $[1,+\infty)$.

In conclusion, the function $f(x)$ has exactly $2$ zero points when $1\leqslant a < 2$ or $a\geqslant 4$.

Therefore, the answer is $\boxed{1\leqslant a < 2 \text{ or } a\geqslant 4}$.

This problem tests our understanding of piecewise functions and the ability to classify and discuss the problem from different angles, breaking down the problem and studying it from multiple perspectives.