When the weight was hung onto a vertical string, the tension in the string was only equal to the weight.
However, when the weight was hung on a horizontal string, the horizontal string cannot remain horizontal. This is because a horizontal string cannot provide any upward component of tension to balance the downward pull of the weight.
At equilibrium, the string makes an angle θ with the vertical.
Since the vertical forces at point X should balance, 2Tcosθ = mg.
Do you see that the tension in the string is actually larger than the weight?
At point Y, we know that T = Mg. so we can work out that 2Mgcosθ = mg, which leads to the result that cosθ = m/(2M). In this video where m/M is 0.5, θ works out to be 75°.