When you put a zinc rod in a acid solution containing $\ce{H^+}$ ions, the zinc piece loses $\ce{Zn^{2+}}$ ions to the solution leaving electrons behind. These electrons attract hydrogen atoms to the zinc rod, and an electron transfer occurs, resulting in bubbles of hydrogen forming on the surface of the zinc. However, the evolution of gaseous hydrogen from zinc surface is rather slow, and that is why pure zinc reacts slowly with dilute acids.
The situation changes when a copper,or any other metal less electropositive than hydrogen is attached to zinc, by a wire. The copper can now be considered as an extension of the zinc rod: a place form where hydrogen can take electrons quickly and become gas. As the rate of evolution of hydrogen from the copper piece is faster, hydrogen bubbles out from the copper piece. This does not mean, however, that the zinc stops reacting with acid; it does, but at a very slow rate.
A common mistake is to think that the copper rod participates in the electrochemical reaction taking place. The copper is merely a pathway to release electrons to H+ ions in the solution. The only reaction taking place is
$\ce{Zn + 2H^+ -> Zn^{2+} +H2\uparrow}$
Copper has no place in the reaction.
So, to answer your questions,
Why does this happen?
This happens because copper can do away with the electrons quicker than zinc, and when electrons get a pathway, they will take the pathway.
Why do some electrons travel through the conductor instead of being attracted immediately by the hydrogen ions in the vinegar?
Well, some electrons do attach to H+ ions from vinegar directly, and hydrogen bubbles should form on the zinc plates, although they are less in amount. And I believe my earlier explanation answers this question.
What if the conductor instead was very long and the pieces of copper and zinc far apart?
It does not have anything to do with distance. If you take a large solution, the reaction will certainly occur. On the other hand, if you separate the poles into two solutions the copper rod will not give off electrons as the circuit is not complete. Think about it this way, if the reaction was to occur, the solution with zinc will become progressively positive and the solution with copper will become negative. This can never happen.
Footnotes:
(1) As per your comment below, I now attempt to answer the question "What makes copper better than zinc in releasing electrons?"
The answer lies in a factor called overpotential, sometimes referred to as overvoltage. There is one type of overpotential which is important in this case— activation overpotential. This is caused due to the fact that when hydrogen atoms are produced on the surfce of a metal ($\ce{H+ +e-> [H]}$), they take some time before they can combine to form diatomic hydrogen and some more time before the hydrogen molecules can be released as gas. This results in the formation of a layer of hydrogen on the surface of metal effectively reducing the rate of reaction and halting it; this effect is known as polarization.
As you can see from the data chart in the Wikipedia page, Zn electrode has -0.77 V overpotential for evolution of hydrogen while there is no overpotential for copper electrode. This means that copper is not polarized easily.
(2) I mentioned in the beginning of my answer that pure zinc reacts with dilute acid slowly. Commercial zinc, on the other hand, reacts quickly with acids, owing of the same effect of polarization. In commercial zinc, impurities of other metals form cathode and the pure zinc forms anode in the acid medium, essentially producing a galvanic cell, where the anode and cathode are connected. Thus the reaction occurs quickly, without the hindrance of polarization.
(3) You should read this paper : http://pubs.acs.org/doi/abs/10.1021/ed078p516. It has some facts which you may find interesting.
