I am not sure if I have computed the below expectations right since it is conditional on starting on 0.
Question:
Let $\left(S_{n}\right)_{n \geq 0}$ be a simple symmetric random walk. Suppose that $m, n, x, y$ are integers such that $n>m>0$, and $n, y$ have the same parity.
Compute $E_{0}\left(S_{m}+S_{n-m} \mid S_{n}=y\right)$ and $E_{0}\left(S_{m} \mid S_{2 m}=2 x\right)$
My attempt:
By the linearity of expectation, I can write $E_{0}\left(S_{m}+S_{n-m} \mid S_{n}=y\right)$ = $E_{0}\left(S_{m} \mid S_{n}=y\right)+E_{0}\left(S_{n-m} \mid S_{n}=y\right)$
$$E_{0}\left(S_{m} \mid S_{n}=y\right)=\sum_{i=1}^{m} P\left(\xi_{i} \mid S_{n}=y\right)=\frac{m y}{n}$$
Similarly,
$$E_{0}\left(S_{n-m} \mid S_{n}=y\right)=\sum_{i=m}^{n} P\left(\xi_{i} \mid S_{n}=y\right)=\frac{(n-m) y}{n}$$
Therefore, $$E_{0}\left(S_{m}+S_{n-m} \mid S_{n}=y\right) = \frac{m y}{n} + \frac{(n-m) y}{n} = y$$
Intuitively this makes sense to me as $S_{m}+S_{n-m}$ should equal to $S_{n}$, therefore the expectation of $E_{0}\left(S_{m}+S_{n-m} \mid S_{n}=y\right) = y$
Similarly, for $E_{0}\left(S_{m} \mid S_{2 m}=2 x\right)$ I got:
$$E_{0}\left(S_{m} \mid S_{2 m}=2 x\right)=\frac{m 2 x}{2 m}=x$$
Again intuitively this makes sense to me as by time $m$, $S_{m} = X$ or else by time $2m$ it will not reach $2x$.
What you did is globally right. However, you have to be a bit more careful in the step $$ E_{0}\left(S_{n-m} \mid S_{n}=y\right)=\sum_{i=m}^{n} P\left(\xi_{i} \mid S_{n}=y\right).$$ Indeed, it is not true that $S_{n-m}$ equals $S_n-S_m$ (actually, $S_{n-m}$ is independent of $S_n-S_m=\sum_{i=n-m+1}^n\xi_i$. However, $S_{n-m}=\sum_{i=1}^{n-m}\xi_i$ hence we actually have $$ E_{0}\left(S_{n-m} \mid S_{n}=y\right)=\sum_{i=1}^{n-m} P\left(\xi_{i} \mid S_{n}=y\right).$$ As you have already observed, $ P\left(\xi_{i} \mid S_{n}=y\right)=y/n$ hence $E_{0}\left(S_{n-m} \mid S_{n}=y\right)=(n-m)y/n$.