How do I solve this probability problem with Probability?

I think you are mixing up the "coordinate" of the breaking point with the length of the resulting stick segment. The length of the resulting stick segment is y-x, (if y>x) not y. If we want x and y to be uniformly distributed, it is is useful to think in terms of the length of the leftmost stick segment l1=Min[x,y] and the length of the center stick segment which is l2=Max[x,y]-Min[x,y] (=Abs[x-y])

This seems to be what you want

Block[{t1, t2, t3},
 t1 = {1 > x > 0, 1 > y > 0};
 t2 = {Min[x, y] + (Max[x, y] - Min[x, y]) > z, 
    Min[x, y] + z > (Max[x, y] - Min[x, y]), (Max[x, y] - Min[x, y]) +
       z > Min[x, y]} /. z -> 1 - Max[x, y];
 Print[t2];
 Print[t3 = And @@ t1~Join~t2 // FullSimplify];
 Probability[t3, 
  Distributed[x, UniformDistribution[{0, 1}]] && 
   Distributed[y, UniformDistribution[{0, 1}]]]]

-> 1/4

Possibly the following is nicer

Block[
 {t1, t2, t3, x, y, l1, l2, l3},
 t1 = {1 > x > 0, 1 > y > 0};
 t2 = {l1 + l2 > l3, l1 + l3 > l2, l2 + l3 > l1} /. {l1 -> Min[x, y], 
    l2 -> Abs[x - y], l3 -> 1 - Max[x, y]};
 Print[t3 = And @@ t1~Join~t2 // FullSimplify];
 Probability[t3, 
  Distributed[x, UniformDistribution[{0, 1}]] && 
   Distributed[y, UniformDistribution[{0, 1}]]]]

-> 1/4

Amazingly, the most direct and mindless possible approach works instantaneously.

Start by characterizing the valid side lengths: each side is the shortest distance between its endpoints; the path made by the other two sides cannot be any shorter.

triangleQ[x_, y_, z_] := x <= y + z && y <= z + x && z <= x + y;

Break the stick uniformly and independently at locations $u_1$ and $u_2$:

f = UniformDistribution[{{0, 1}, {0, 1}}];

Noting that the breaks (from left to right) create pieces of length $\min(u_1,u_2)$, $|u_2-u_1|$, and $1 - \max(u_1, u_2)$, request the probability that the pieces form a triangle:

Probability[triangleQ[Min[u1, u2], Abs[u2 - u1], 1 - Max[u1, u2]], {u1, u2} \[Distributed] f ]

$\frac{1}{4}$

Each uniform divides the stick into a smaller and a longer side. If the smaller side of both uniforms coincide (left-left, or right-right) then you won't have a triangle because the rightmost/leftmost division will be longer than 0.5. The odds of this happening is 1/2. If they don't coincide, then you will have a triangle only when the sum of the smallest sides is higher than 0.5. So,

1/2 Probability[
  x + y < 1/2, {x, y} \[Distributed] 
    TransformedDistribution[Min[z, 1 - z], 
     z \[Distributed] UniformDistribution[]] // Thread]

(* 1/4 *)

By the way

With[{min := Min[x, y], max := Max[x, y]}, 
 RegionPlot[
  And @@ Thread[0 < {x, y} < 1] && 
   Max[min, max - min, 1 - max ] < 1/2, {x, 0, 1}, {y, 0, 1}]]