How does zinc become oxidized when it's not in direct contact with copper ions in a galvanic cell? Whenever a galvanic cell is explained in textbooks/videos, it usually starts by showing that if you put a piece of $\ce{Zn}$ in a $\ce{Cu^{2+}}$ ion solution, the $\ce{Zn}$ will become oxidized and transfer $\ce{2e-}$ to the $\ce{Cu^{2+}}$ ion, so the $\ce{Cu^{2+}}$ becomes a neutral $\ce{Cu}$ metal. 

I understand this, but in a galvanic cell, the $\ce{Zn}$ never actually touches the $\ce{Cu^{2+}}$ ions, the $\ce{Zn}$ and $\ce{Cu}$ are only connected with wires. I don't understand how the $\ce{Zn}$ can be oxidized and transfer ions through the current.
Zinc electrode is anode and copper electrode is cathode. The two electrodes are connected by a electrically conducting wire. At anode, $\ce{Zn}$ atoms are oxidised to $\ce{Zn^2+}$ which go into the solution, and the electrons are left on the metal strip. 

The electrons produced in the oxidation half-reaction at anode are transferred through the wire to $\ce{Cu}$ electrode where they are consumed. The $\ce{Cu^2+}$ ions are reduced to metallic $\ce{Cu}$ which coats on the $\ce{Cu}$ strip.

Thus, oxidation and reduction half reactions occur at separate electrodes and the current flows through the wire. A salt bridge maintains neutrality in both the compartments by a flow of ions. 

The overall cell reaction can be given as:

$$\ce{Zn (s) + Cu^2+ -&gt; Zn^2+ + Cu (s)}$$