If metals want to "lose" electrons, then why will copper ions take electrons from a zinc electrode?
Typically, when one uses terms such as "likes to lose" or "likes to gain", we should think of that phrase as "likes to lose relative to...".  In your case, when we think about electron configurations, it might be more helpful to think "metals like to lose rather than gain electrons to achieve a noble-gas-like configuration".  When comparing two metals, both of which tend to lose electrons, we need to look at different concepts to answer the question.

We can explain the transfer of electrons by looking at the the metal activity series such as the one shown here.  The way one reads this table is that a Galvanic cell created with two metals will have the metal higher on the list as its anode (where oxidation occurs) and the other metal will be the cathode.  Thus, we would expect Zn to be oxidized to Zn(II) and Cu(II) to be reduced to Cu.  These tables are based on standard reduction potentials: 

$$\ce{Cu^{2+} + 2 e^- -&gt; Cu(s)}\ \tag{E = 0.34 V}$$
$$\ce{Zn^{2+} + 2 e^- -&gt; Zn(s)}\ \tag{E = -0.76  V}$$

By flipping the 2nd reaction (thus changing the sign of E) one obtains a cell potential of about 1.1 V; a positive number meaning that this reaction will happen spontaneously:

$$\ce{Cu^{2+} + Zn(s) -&gt; Cu(s) + Zn^{2+}}\tag{E = 1.1V}$$

The standard reduction potentials tell us that the electrons would rather be on copper than on zinc.