I'd like to put forth an answer which directly addresses the title of your post, but not the particular situation in which you put forth with the meter stick and rope.
Consider instead a massive rope hanging vertically from a ceiling.
Give the rope a total mass of, say, $M$. Then use Newton's second law on the lower half of the rope to find the tension at the midpoint. Compare this value to the tension at the top of the rope by using Newton's second law for the entire rope. This should let you answer your question.
Say you have a weight tied to each side a a rope which is strung over a pulley with friction. Here's a really easy way to see why the tensions on each side of the rope can't be equal.
Imagine a really stiff pulley - in other words, ${\bf F}_\text{friction}$ is high. If that's the case, it'll be possible to balance unequal loads on this pulley system - i.e. a heavy weight on the right side and a lighter weight on th left - without the system moving. If the weights don't move, then we can say that the forces acting on each weight add up to zero:
For the heavy weight, there's the weight downward, ${\bf w}_\text{heavy}$ and there's the tension of the right side of the rope upward, ${\bf T}_\text{right}$. The tension pulls up and the weight down, and the system doesn't move, so
$$ {\bf T}_\text{right} - {\bf w}_\text{heavy} = 0
$$
or
$$ {\bf T}_\text{right} = {\bf w}_\text{heavy}
$$
Similarly for the left (light) side,
$$ {\bf T}_\text{left} - {\bf w}_\text{light} = 0 \quad \Rightarrow \quad{\bf T}_\text{left} = {\bf w}_\text{light}
$$
As you can see, the tension on the right, ${\bf T}_\text{right}$ is equal in magnitude to the heavy weight, while the tension on the left, ${\bf T}_\text{left}$ is equal to that of the lighter weight. The friction is introducing an extra force which changes the tensions on each side.
As far as your question about rope stretching goes, if you anchor a rope on one side and pull, the rope will pull back, creating a tension. This is indeed because of stretching in the rope. This is not really what Newton's 3rd law is referring to. Newton's third law, in this case, tells us that the force that we feel from the rope, tension, is exactly the force the rope feels from us pulling. The two are equal and opposite. You can change the tension by changing the stiffness of the rope, but whatever the tension, Newton's 3rd law will still be true - the rope will feel us pulling it as much as we feel it pulling us.
A mass m hangs on 3 ropes symmetrically with equal base angles.
Best Answer
Let the vertical distance between the ceiling and the point where the 3 ropes meet be $L+\Delta L$. The original length of the side rope is $L\sqrt{2}$. From the pythagorean theorem, the new length of the side rope is $\sqrt{L^2+(L+\Delta L)^2}=\sqrt{2L^2+2L\Delta L+(\Delta L)^2}$. If we linearize this with respect to $\Delta L$, we obtain $L\sqrt{2}(1+\frac{\Delta L}{2L})$. So the change in length of the side rope is $$L\sqrt{2}(1+\frac{\Delta L}{2L})-L\sqrt{2}=\frac{1}{\sqrt{2}}\Delta L$$So the tensile strain in the side rope is $$\epsilon=\frac{(L/\sqrt{2})}{(L\sqrt{2})}=\frac{1}{2}\frac{\Delta L}{L}$$So the tension in the side rope is $$T_{SR}=\frac{EA}{2}\frac{\Delta L}{L}$$ The tension in the vertical rope segment between the ceiling and the point where the three ropes meet is: $$T_{VR}=EA\frac{\Delta L}{L}$$ So the tension in the side rope is half the tension in the vertical rope segment.
The remainder of the analysis is straightforward, and just involves doing the equilibrium force balance.