Let $\{N(t), t\geq 0\}$ be a Poisson process with rate $\lambda$, $S_n$ the instant of the $n$-th arrival and $T_n$ the $n$-th interarrival time, that is, $T_n = S_n - S_{n-1}$, $n \geq 1$.
Now consider the following result:
Theorem. Given that $N(t) = n$, the $n$ arrival times $S_1, S_2, \dots, S_n$ have the same distribution as the order statistics corresponding to $n$ independent random variables uniformly distributed on the interval $(0,t)$.
I would like to know how to calculate $\mathbb{E}[S_4 | N(1) = 2]$ using the theorem above.
I have already solved it using the memorylessness property of the exponential distribution, since $T_i \sim Exponential(\lambda)$, and it went like: you can call $S_4 = 1 + T_3 + T_4$, then \begin{align*} \mathbb{E}[S_4 | N(1) = 2] &= \mathbb{E}[1 + T_3 + T_4] \\ &= 1 + \mathbb{E}[T_3 + T_4] \\ &= 1 + \frac{2}{\lambda}, \end{align*} since $(T_3 + T_4) \sim Gamma(2, \lambda)$, so I know the result I should get.
My attempt: we can write $S_4 = (T_1 + T_2) + T_3 + T_4 = S_2 + T_3 + T_4$, so it follows \begin{align*} \mathbb{E}[S_4 | N(1) = 2] &= \mathbb{E}[S_2 + T_3 + T_4 | N(1) = 2] \\ &= \mathbb{E}[S_2 | N(1) = 2] + \mathbb{E}[T_3 + T_4] \\ &= \frac{1}{2} + \frac{2}{\lambda}, \end{align*} since increments are independent, $(S_2|N(1) = 2) = \max \{U_1, U_2\}, U_i \sim Uniform(0,1)$, and $(T_3 + T_4) \sim Gamma(2, \lambda)$.
What have I done incorrectly? I'd get the correct result if I wrote instead $S_4' = S_1 + S_2 + T_3 + T_4$, but that's absurd since \begin{align*} S_4' &= S_1 + S_2 + T_3 + T_4 \\ &= (T_1) + (T_1 + T_2) + T_3 + T_4 \\ &= T_1 + (T_1 + T_2 + T_3 + T_4) \\ &= T_1 + S_4. \end{align*}
In addition to that, in these MIT freely available online class notes from a "Discrete Stochastic Processes" course, on page $92$, we have equation $(2.46)$: \begin{align} \mathbb{E}[S_i|N(t) = n] = \frac{it}{n+1}, \end{align} which in my attempt would yield a completely different result: \begin{align} \mathbb{E}[S_2|N(1) = 2] = \frac{2}{3}. \end{align}
How to proceed?