I think the best way to see this is through the graph and a small example.
For the graph:
http://www.engineeringtoolbox.com/arithmetic-logarithmic-mean-temperature-d_436.html
Just looking at the graph, notice how the counter flow will continuously cool the wort because it is exposed to colder and colder water as it flows through. Essentially, if the heat exchanger were infinitely long (or have an infinite surface) the wort would reach the temperature of the inlet water.
Now compare it to parallel flow: if there were infinite surface or infinitely long, the coolest the wort would get would be somewhere between the water inlet temp and the wort inlet temp. The wort could never get as cold as cold as the inlet water temp because it is exposed to warmer and warmer water as it flows along and less and less transfer occurs. Hence it lands somewhere in between.
Now if we want to compare them using the log mean temperature difference (LMTD) using the generalized form which should be just fine for an example:
LMTD=
dT1-dT2
ln(dT1/dT2)
where dT1 is the difference between the two inlets/outlets on one side, and dT2 is the difference of inlets/oulets of the other side. Assuming water comes in at 10 C and wort at 100 C, and the exiting wort is at 20 C (usually the conditions I have) and leaving the water's outlet temperature is unknown but defined as A for the counterflow and B for parallel and without making any assumptions this gives:
For counterflow:
(10-20)-(A-100)
ln(10-20/1-100)
=
90-A
ln(-10/A-100)
For parallel:
(100-10)-(20-B)
ln(100-10/20-A)
=
70-B
ln(90/20-B)
so we can see right off the bat that the numerator is larger in the counterflow exchanger. If we look at the ln's we see that the smallest value will be generated by the counterflow exchanger as well. This means, twice over, that the counterflow will have a larger LMTD. But what about the rest of the heat transfer equation? Well q=h*pi*D*L*lmtd, and we can say that since it is wort and water in both cases h will not change, and if we are going to compare exchangers I would say we leave D and L the same as well. Of course there is a way to adjust D and L until you have the same result as counterflow, but that is both expensive and hard on your equipment (hottest fluid meeting coldest fluid continuously creates harsh thermal expansions and contractions on the metals).
So to try and say this simply, parallel introduces the wort to colder and colder water the further it travels. Parallel introduces the wort to cold water right away but then the same water for the rest of the trip which gets warmer and warmer.
Just to return to the graphs and address the other point where they mention that the whole trip needs to be taken into account, let's use enthalpy and integration (somewhat loosely). The total heat transfer will be equal to the sum of the enthalpy across all points. The enthalpy at any point will (considering it's the same fluids, lengths, and so on and on) let's say be the temperature difference. That means the total transfer is the integral of the wort minus the integral of the water. Now we can go back to simple terms. It's the surface area between the lines on the graph. The surface will be large at first with the parallel flow, and continuously get smaller quite rapidly. The counterflow will be smaller at first but not for long as it will be larger than the parallel who is shrinking more rapidly. Then the counterflow will have a greater surface. This only works as an analysis if the exchanger is long enough to allow this to happen. If it is ridiculously short (you won't get to fermenting temperature if it is), then yes parallel would be better but both would be useless for brewing or anything else that requires more than a couple degrees of cooling/heating. So counterflow will be more efficient graphically - as long as we stay within reason - as well.
I hope this helps, cheers.
