Why does it yield iron(II) chloride instead of iron when copper meets iron(III) chloride?
Take a look at standard reduction potentials at $\pu{25 ^\circ C}$:

$$
\begin{array}{clr}
\hline
\text{Reaction} &amp; E^\circ~(\pu{V}) \\
\hline
  \begin{align}
    \ce{Fe^3+ + e- &amp;-&gt; Fe^2+} \\
    \ce{Cu^2+ + 2e- &amp;-&gt; Cu} \\
    \ce{Fe^3+ + 3e- &amp;-&gt; Fe} \\
    \ce{Fe^2+ + 2e- &amp;-&gt; Fe}
  \end{align}
&amp;
  \begin{array}{r}
    +0.77 \\
    +0.34 \\
    -0.04 \\
    -0.41
  \end{array} \\
\hline
\end{array}
$$

A more positive potential means a greater tendency to be reduced, so $\ce{Cu(II)}$ is easier to reduce to copper metal than either $\ce{Fe(II)}$ or $\ce{Fe(III)}$ being reduced to iron metal. So, copper metal can't force the formation of iron metal.

But $\ce{Fe(III)}$ can be reduced just to $\ce{Fe(II)}$ at a relatively high potential, high enough to allow the copper to be in its oxidized form as $\ce{Cu(II)}$. So, we can have $\ce{Cu(II)}$ and $\ce{Fe(II)}$ at equilibrium and thus we can get the corresponding chlorides $\ce{CuCl2}$, $\ce{FeCl2}$.