$U \sim U(0,1)$
Define $X = \text{max}\{U, 1-U\}$
so that
$$X =
\begin{cases}
1-U, & 0 < U < 1/2 \\
U, & 1/2 \leq U < 1
\end{cases}
$$
It follows that
$$X = \frac{1 + |2U - 1|}{2}$$
$$F_X(x) = P (X \leq x), \:\:\:\: 1/2 < x < 1$$
$$=P\left(\frac{1 + |2U - 1|}{2} \leq x \right)$$
$$=P(|2U - 1| \leq 2x-1)$$
$$=P(-2x+1 \leq 2U - 1 \leq 2x-1)$$
$$=P(1-x \leq U \leq x)$$
$$=P(1-x < U \leq x)$$
$$=F_U(x) - F_U(1-x)$$
Differentiating with respect to $x$:
$$f_X(x) = f_U(x) + f_U(1-x), \:\:\:\:\: \text{for}\:\:\: 1/2 < x < 1$$
It follows that
$$f_X(x) =
\begin{cases}
2, & \text{if}\:\:\:1/2 < x < 1, \\
0, & \text{otherwise}
\end{cases}
$$
and since we have for a uniform variable
$$f_Z(z) =
\begin{cases}
\frac{1}{b-a}, & \text{if}\:\:\:a < z < b, \\
0, & \text{otherwise}
\end{cases}
$$
$$E[Z] = \frac{b+a}{2} $$
It follows that
$$E[X] = \frac{1 + 1/2}{2} = \frac{3}{4} $$
Your problem is that what you have obtained (the ratio of hatched triangle's area to big triangle area) is
$$P(X\color{red}{\ge s})=(1-3s)^2$$
Therefore :
$$F(s):=P(X<s)=1-P(X \geq s)=1-(1-3s)^2$$
Deriving it, you will get :
$$f(s)=6(1-3s) \ \ for \ 0 \leq s \leq \frac13$$
from which the mean is easily obtained.
Best Answer
$U$ is the break point which lies uniformly distributed on $(0;1)$.
$L$ is the length of the longer side of the break. This is somewhere on $[\tfrac 1 2; 1)$ .
When the break point $U$ is less than $1/2$ the length of the longer stick is $1-U$, and other wise it is $U$.
So, for any $\tfrac 1 2\leq l\leq 1$, then the longer length being less than $l$ means that the break is both less that $l$ and greater than $1-l$ .
$$\mathsf P(L<l) ~=~ \mathsf P(1-l<U<l) \\ = ~~\,2l-1 \\ = ~ \frac{l-\tfrac 12}{1-\tfrac 12} $$
Which means $L\sim\mathcal {U}(\tfrac 12;1)$