$\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\eD}{\overset{\text{D}}\to}
\newcommand{\D}{\overset{\text{D}}=}$
Let $U_1,U_2,\dots$ be iid random variables, each uniformly distributed on $[0,1]$. For a fixed natural $k$, let $Y_{n,k}$ be the $k$th largest value among $U_1,\dots,U_n$. For a fixed $p\in(0,1)$, let $q_{n,k}(p)$ be the $p$-quantile of $Y_{n,k}$. The problem then is to find
\begin{equation}
\lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p)).
\end{equation}
Note that $Y_{n,k}$ has the beta distribution with parameters $n+1-k,k$. So (see Sections Derived_from other distributions and Summation),
\begin{equation}
1-Y_{n,k}\D\frac{S_k}{S_{n+1}},
\end{equation}
where $\D$ denotes the equality in distribution, $S_j:=X_1+\dots+X_j$, and $X_1,X_2,\dots$ are iid standard exponential r.v.'s. So, by the law of large numbers,
\begin{equation}
S_{n,k}:=n(1-Y_{n,k})\eD S_k,
\end{equation}
where $\eD$ denotes the convergence in distribution.
(Note also that $S_k$ has the gamma distribution with parameters $k$ and $1$.)
So, $n(1-q_{n,k}(p))=\tilde q_{n,k}(1-p)\to \tilde q_k(1-p)$, where $\tilde q_{n,k}(1-p)$ and $\tilde q_k(1-p)$ denote the $(1-p)$-quantiles of $S_{n,k}$ and $S_k$, respectively. Hence,
\begin{align}
n(1-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p))
&=\E(S_{n,k}|Y_{n,k}\le q_{n,k}(p)) \\
&=\E(S_{n,k}|S_{n,k}\ge \tilde q_{n,k}(1-p)) \\
&\to\E(S_k|S_k\ge \tilde q_k(1-p)),
\end{align}
by an appropriate uniform integrability involving (say) second moments.
Thus,
\begin{equation}
\lim_{n\to\infty}n(q_{n,k}(p)-\E(Y_{n,k}|Y_{n,k}\le q_{n,k}(p))
=\E(S_k|S_k\ge \tilde q_k(1-p))-\tilde q_k(1-p).
\end{equation}
