$\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\D}{\overset{D}=}$Let $X_{n:1},\dots,X_{n:n}$ denote the order statistics in question. Fix any real $t>0$. You wanted to show that for natural $k\le n$
\begin{equation*}
	p_{n,k}:=P\Big(\sum_{j=1}^k X_{n:j}
	\ge(1+t)\frac{k(k+1)}{2(n+1)}\Big)\to0 \tag{10}\label{10}
\end{equation*}
as $n\to\infty$. As [noted by James Martin][1], \eqref{10} will fail to hold unless $k\to\infty$. 

So, we henceforth assume that $k\to\infty$ (and hence $n\to\infty$), and then we will show that \eqref{10} holds. Moreover, we will obtain an explicit upper bound on $p_{n,k}$, which goes to $0$ exponentially fast with $k$. 

First here, consider the gaps 
\begin{equation*}
	G_i:=X_{n:i}-X_{n:i-1}
\end{equation*}
between the order statistics for $i=1,\dots,n+1$, where $X_{n:0}:=0$ and $X_{n:n+1}:=1$, so that 
\begin{equation*}
	X_{n:j}=\sum_{i=1}^j G_i
\end{equation*}
for $j=1,\dots,n$. 

Next, recall that 
\begin{equation*}
	(G_1,\dots,G_{n+1})\D\frac{(Y_1,\dots,Y_{n+1})}{Y_1+\dots+Y_{n+1}}, 
\end{equation*}
where $\D$ means the equality in distribution and $Y_1,\dots,Y_{n+1}$ are independent random variables each with the exponential distribution with mean $1$. So, 
\begin{equation*}
	\sum_{j=1}^k X_{n:j}=\sum_{j=1}^k \sum_{i=1}^j G_i
	=\sum_{i=1}^k (k-i+1)G_i\D\sum_{i=1}^k iG_i \\ 
	\D\frac{\sum_{i=1}^k iY_i}{\sum_{i=1}^{n+1}Y_i}. 
\end{equation*}
So, taking a real $\ep>0$ small enough so that 
\begin{equation*}
	\de:=(1+t)(1-\ep)-1>0, \tag{15}\label{15}
\end{equation*}
we see that 
\begin{equation*}
	p_{n,k}=P\Big(\frac{\sum_{i=1}^k iY_i}{\sum_{i=1}^{n+1}Y_i}
	\ge(1+t)\frac{k(k+1)}{2(n+1)}\Big)\le p_n+q_k, \tag{20}\label{20}
\end{equation*}
where 
\begin{equation*}
	p_n:=P\Big(\sum_{i=1}^{n+1}Y_i<(1-\ep)(n+1)\Big),
\end{equation*}
\begin{equation*}
	q_k:=P\Big(\sum_{i=1}^k iY_i\ge(1+\de)\frac{k(k+1)}{2}\Big). 
\end{equation*}
Further, for any real $h\in(0,1)$, by the Bernstein--Chernoff inequality,
\begin{equation*}
	p_n=P\Big(\sum_{i=1}^{n+1}(1-Y_i)>\ep(n+1)\Big) \\ 
	\le e^{-h\ep(n+1)}E\exp h\sum_{i=1}^{n+1}(1-Y_i)
	=\Big(\frac{e^{h(1-\ep)}}{1+h}\Big)^{n+1}.
\end{equation*}
The latter expression is minimized in $h$ when $h=\ep/(1-\ep)$, and then we get 
\begin{equation*}
	p_n\le p_{n,\ep}:=\big((1-\ep)e^\ep\big)^{n+1}; \tag{30}\label{30}
\end{equation*}
note that $p_{n,\ep}\to0$ exponentially fast with $n$, because $(1-\ep)e^\ep<1$. 

Further yet, for any real $h\in(0,1/k)$, again by the Bernstein--Chernoff inequality,
\begin{equation*}
	q_k\le\exp\Big\{-h(1+\de)\frac{k(k+1)}{2}\Big\}
	E\exp h\sum_{i=1}^k iY_i \\ 
	=\exp\Big\{-h(1+\de)\frac{k(k+1)}{2}\Big\}\prod_{i=1}^k\frac1{1-hi}. 
\end{equation*}
Note now that $g(u):=\frac1u\,\ln\frac1{1-u}$ is increasing in $u\in(0,1)$, from $g(0+)=1$. So, 
\begin{equation*}
	\frac1{1-u}=e^{g(u)u}\le e^{g(u_0)u} \tag{35}\label{35}
\end{equation*}
for $u\in(0,u_0]$, and 
\begin{equation*}
	g(u_0)<1+\de
\end{equation*}
if $u_0>0$ is small enough. So, letting now $h=u_0/k$ and using \eqref{35} with $u=hi$, we get 
\begin{equation*}
	q_k\le q_{k,\ep,u_0}
	:=\exp\Big\{-u_0(1+\de-g(u_0))\frac{k+1}{2}\Big\}, \tag{40}\label{40}
\end{equation*}
and $q_{k,\ep,u_0}\to0$ exponentially fast with $k$.  

Collecting \eqref{20}, \eqref{30}, and \eqref{40}, we get 
\begin{equation*}
	p_{n,k}\le p_{n,\ep}+q_{k,\ep,u_0},
\end{equation*}
and the latter upper bound on $p_{n,k}$ converges to $0$ exponentially fast with $k$. The optimal choice of the parameters $\ep,u_0$ (with $\de$ defined in \eqref{15}) depends on the value of $t>0$ in \eqref{10} and on how fast $k$ grows in relation with $n$. 

[1]: https://mathoverflow.net/questions/463406/concentration-for-sum-of-order-statistics#comment1203138_463406