The three triangle inequalities are
\begin{align}
x + y &> 1-x-y \\
x + (1-x-y) &> y \\
y + (1-x-y) &> x \\
\end{align}
Your problem is that in picking the smaller number first from a uniform distribution, it's going to end up being bigger than it would if you had just picked two random numbers and taken the smaller one. (You'll end up with an average value of $1/2$ for the smaller instead of $1/3$ like you actually want.) Now when you pick $y$ on $[0, 1-x]$, you're making it smaller than it should be (ending up with average value of $1/4$). To understand this unequal distribution, we can substitute $y (1-x)$ for $y$ in the original inequalities and we'll see the proper distribution.
(Note that the $y$-axis of the graph doesn't really go from $0$ to $1$; instead the top represents the line $y=1-x$. I'm showing it as a square because that's how the probabilities you were calculating were being generated.) Now the probability you're measuring is the area of the strangely-shaped region on the left, which is
$$\int_0^{1/2}\frac1{2-2x}-\frac{2x-1}{2x-2}\,dx=\ln 2-\frac12\approx0.19314$$
I believe that's the answer you calculated.
Not an answer, but it might help to progress further. [The derivation that follows provides a strict -but practically useless- bound. The second part of the answer has some results that might be of interest, but they are merely empirical]
Let's consider the (much more restricted) event that the six lengths form a tetrahedron in whatever order. In the linked pdf this set of lengths is called "completely tetrahedral", and a necessary-sufficient condition is given (Theorem 4.2) which is equivalent to the following: $ u \le \sqrt{2}v $, where $u,v$ are the maximum and minimum lengths. This, informally, would correspond to "almost regular" tetrahedra.
Let's then compute the probability that the lengths are completely tetrahedral. Because the points are chosen at random, uniformly, the lengths are probabilistically equivalent to a set of iid exponential variables with arbitrary parameter, conditioned to a constant sum. Because we are only interested in ratios, we can even ignore this conditioning.
Now, the joint probability of the maximum and minimum of a set of $n$ iid variables is given by
$$ f_{u,v}= n(n-1) f(u) f(v) [F(u) -F(v)]^{n-2}, \hskip{1cm} u\ge v$$
In our case: $n=6$, $f(u)=e^{-u}$, and the probability that $u<a \, v$ is a straightforward integral, which gives:
$$P(u<a \, v)= 5 (a -1)\left( \frac{1}{1+5\,a}-\frac{4}{2+4\,a}+\frac{6}{3+3\,a}-\frac{4}{4+2\,a}+\frac{1}{5+1\,a} \right)$$
And $P(u<\sqrt{2} v) \approx 7.46 \times 10^{-5}$
This should be a strict bound on the desired probability, but, surely, far from tight.
[Update: indeed, the bound is indeed practically useless, it corresponds to an irrelevant tail. The probability, as per my simulations, is around $p=0.06528$ ($N=10^9$ tries, $3 \, \sigma \approx 2.3 \times 10^{-5}$), which agrees with other results.]
The only empirical result that might be of interest: It's easy to see that, from the $6!$ possible permutations of the lenghts, we can restrict ourselves to $30$, from symmetry considerations; now, from my simulations, I've found that it's sufficient to consider 7 permutations, the first two being already enough for more than $90\%$ of the successes;
and the (need to consider the) seventh one is extremely small. These permutations are:
$$p_1 = [0 , 1 , 4 , 5 , 3 , 2] \hskip{1 cm} (0.75837)\\
p_2 = [0 , 1 , 4 , 3 , 5 , 2] \hskip{1 cm} (0.15231)\\
p_3 = [0 , 2 , 4 , 1 , 5 , 3] \hskip{1 cm} (0.08165)\\
p_4 = [0 , 1 , 4 , 5 , 2 , 3] \hskip{1 cm} (0.00404)\\
p_5 = [0 , 1 , 4 , 2 , 5 , 3] \hskip{1 cm} (0.00245)\\
p_6 = [0 , 1 , 3 , 5 , 4 , 2] \hskip{1 cm} (0.00116)\\
p_7 = [0 , 1 , 3 , 4 , 5 , 2] \hskip{1 cm} (0.00002)\\
$$
The length indexes correspond to a sorted array (say, ascending), and following the convention of the linked paper: the first three sides have a common vertex, the following three are the corresponding opposite sides (so, for example, in the first permutation, and by far the most favorable one, the longest and shortest sides are opposite). The numbers on the right are the probability that this permutation (when one tries in the above order) is the successful one (given that they form a tetrahedron). I cannot be totally sure if there is some rare case that requires other permutation (very improbable, I'd say), but I'm quite sure (unless I've made some mistake) that the set cannot be further reduced.
Best Answer
The probability is exactly
$$\frac{5^2\cdot 29^2}{2^2\cdot 7\cdot 11\cdot 13\cdot 19}=\frac{21025}{76076}\approx 0.2763683684736316$$
We're trying to compute the volume of a rather complicated region in 5D space, but one which is ultimately bounded by facets with rational coordinates, so we know the answer will have to be rational in the end. We just want to find a way of cutting up this big 5D polytope inside the unit cube into pieces we can integrate.
The first simplification we'll make is to assume that the five cuts are ordered smallest to largest, without loss of generality, i.e. $0\le x_1\le x_2\ldots\le x_5\le 1$. This means we'll need an extra factor of $120$ at the end, but we can now talk about the lengths of the six segments as $x_1, x_2-x_1, \ldots, x_5-x_4, 1-x_5$.
In this framing, every possible way to arrange the segments into triangles corresponds to a set of six linear inequalities, one for each side of each triangle asserting that that side is at least as large as the sum of the other two sides on its triangle.
So for any given set of ways we can arrange the segments into two triangles, we'll have a convex region described by some half-spaces corresponding to the breakpoints in the large segment such that we can make the pieces into triangles in all of those ways (e.g. we might ask about the splits such that segments (1,2,3) and (4,5,6) make triangles, as well as segments (1,4,5) and (2,3,6)).
Ultimately, we want to know the volume of the region for which at least one of these $\frac{{}_6C_3}2=10$ triangle pairs are possible, so we'll use some inclusion-exclusion counting:
$$\text{Vol}(\text{at least one region}) = \sum_{\emptyset\neq S\subset \{1,2,\ldots,10\}}(-1)^{|S|+1}\text{Vol}\left(\bigcap_{i\in S} \text{triangle pair $i$ is possible}\right)$$
For each of these $1023$ regions* we want to compute, we just need to compute the volume of a particular region in 5-space bounded by a set of rationally-parametrized half spaces. This can be done numerically in Python using
scipy.spatial.HalfspaceIntersectionandscipy.spatial.ConvexHull. To convert a floating-point volume $V$ to an exact value, I looked at the coordinates of the intersections of the region, found a scaling factor $F$ that made those intersections lie at integer points, and then searched for small rational values $q$ such that $qFV$ was very close to an integer. In all cases there was a rational value within a factor of $10^{-14}$ of the floating-point approximation whose denominator had no prime factors over $19$. Then I just used exact integer arithmetic on the resulting fractions to obtain the final answer. (As supporting evidence for these conversions to rational numbers, one of these regions had a volume of $2491157/41409688320$, which is much more complicated than the final sum but which one wouldn't expect to cancel out if it were a random coincidence.)Given how comparatively nice the final answer ends up being, I suspect there is some deeper reason that things work out so well; the numerator and denominator are both suspiciously nice compared to the volumes of the intermediate regions in this sum. I have no idea what approaches might shed light on the reasons behind this, though.
*Because of the symmetry between the six segments we break the stick into, many of these regions are congruent - we only end up needing to consider 18 different possibilities.