ω is a set could be proved in ZF system by axiom of infinity and axiom of separation. Could 2ω, that is, ω + n when n tends to infinite, is a set be proved in ZF system?
The axiom of infinity guarantees that there is at least one inductive set. But it doesn't guarantee that there are more than one inductive set. ω is an inductive set. Then the axiom of infinity cannot guarantee that there is other inductive set.
First of all, $2\cdot\omega=\omega$, and $\omega+\omega=\omega\cdot2$. It's confusing, yes.
Secondly, yes, $\sf ZF$ can prove that $\omega+\omega$ is a set because it has the Replacement axiom schema. This means that if we define a function (e.g. mapping each element of $\omega$ to $\omega+n$), then the range of that function---when restricted to a set---is also a set.
More to the point, $\varphi(x,y)$ is defined as if $x\in\omega$, then $y=\omega+n$ defines a function when considering the domain $\omega$. Therefore the range of $\varphi$ when applied to $\omega$ is itself a set. This is $\{\omega+n\mid n\in\omega\}$. Taking the union of this set will give us exactly $\omega+\omega$, and I'll leave you to figure out why is that.
Without Replacement, though, you cannot prove the existence of $\omega\cdot2$, because $V_{\omega+\omega}$ satisfies all the axioms of $\sf ZF$ except Replacement, and there $\omega+\omega$ does not exist.