Let $X_{1},X_{2},\ldots,X_{m},\ldots,X_{n}$ be independently drawn from a distribution $F$ and let $Y_{k}^{(n)}$ be the $k$-th order statistic (Convention: $Y_{1}^{(n)}>Y_{2}^{(n)}>\cdots>Y_{n}^{(n)}$). What is the probability distribution of the maximum of $Y_{1}^{(n)}$ and $Y_{2}^{(m)}+a$ with $a$ some positive constant. Note that the first order statistic is taken over the whole $n$ draws, while the second order statistic is taken over the subsample of $m$ draws.
So how is $\max\{Y_1^{(n)},Y_2^{(m)}+a\}$ distributed? I think there are 2 possibilities:
The first order statistic of $n$ is not in the $m$ group, then we look at the maximum between the highest draw out of $n-m$ and the second highest draw out of $m$ plus $a$. These two would be independent, so the maximum would just be the product of the two cdfs., \begin{equation} P\big\{\max\{Y_1^{(n-m)},Y_2^{(m)}+a\}\big\}=F^{(n-m)}_{1}(x)F^{(m)}_{2}(x-a) \end{equation}
The first order statistic of $n$ is in the $m$ group, then we look at the maximum between the highest draw out of $m$ and the second highest draw out of $m$ plus $a$, which is from an earlier question here on SE given by \begin{equation} P\big\{\max\left \{Y_{1}^{(m)},Y_{2}^{(m)}+a\right \}<x\big\}=mF(x-a)^{m-1}F(x)-(m-1)F(x-a)^{m} \end{equation}
With this reasoning the probability distribution of $\max\{Y_1^{(n)},Y_2^{(m)}+a\}$ would be given by \begin{gather} P\big\{Y_{1}^{(n)}\notin \text{Group m} \big\}P\big\{\max\{Y_{1}^{(n-m)},Y_{2}^{(m)}+a\}<x\big\} \\ + P\big\{Y_{1}^{(n)}\in \text{Group m} \big\}P\big\{\max\{Y_{1}^{(m)},Y_{2}^{(m)}+a\}<x\big\}\\ =\Big(1-\frac{m}{n}\Big)\Big(\underbrace{F(x)^{n-m}}_{=F^{(n-m)}_{1}(x)}\underbrace{\big(mF(x-a)^{m-1}-(m-1)F(x-a)^{m}\big)}_{=F^{(m)}_{2}(x-a)}\Big)\\ +\frac{m}{n}\Big(\underbrace{mF(x-a)^{m-1}F(x)-(m-1)F(x-a)^{m}}_{=P\big\{\max\left \{Y_{1}^{(m)},Y_{2}^{(m)}+a\right \}<x\big\}}\Big). \end{gather}
Did I miss something?
No, you're ignoring various dependencies. In particular, where you write "These two would be independent", they're not; you're including cases where $Y_1^{(n-m)}\lt Y_2^{(m)}$, despite previously assuming that $Y_1^{(n)}$ is greater than the first $m$ values.
The probability that exactly $k$ of the values are in $(x-a,x]$ and the other $n-k$ are at most $x-a$ is
$$ \binom nkF(x-a)^{n-k}\left(F(x)-F(x-a)\right)^k. $$
The probability that the first $m$ samples are all among the $k$ values or $m-1$ of them are among the $k$ values and one isn't is
$$ \frac{\binom km+(n-k)\binom k{m-1}}{\binom nm}\;, $$
Thus the desired probability is
$$ \binom nm^{-1}\sum_{k=m}^n\left(\binom km+(n-k)\binom k{m-1}\right)\binom nkF(x-a)^{n-k}\left(F(x)-F(x-a)\right)^k. $$