Haskell

It’s more complicated than it needs to be, could’ve done the first part just like the second.
Also it takes one second (!) to run it .-.

import Data.Maybe as Maybe
import Data.List as List
import Control.Arrow hiding (first, second)

parseRule :: String -> (Int, Int)
parseRule s = (read . take 2 &&& read . drop 3) s

replace t r c = if t == c then r else c

parse :: String -> ([(Int, Int)], [[Int]])
parse s = (map parseRule rules, map (map read . words) updates)
        where
                rules = takeWhile (/= "") . lines $ s
                updates = init . map (map (replace ',' ' ')) . drop 1 . dropWhile (/= "") . lines $ s

validRule (pairLeft, pairRight) (ruleLeft, ruleRight)
        | pairLeft == ruleRight && pairRight == ruleLeft = False
        | otherwise = True

validatePair rs p = all (validRule p) rs

validateUpdate rs u = all (validatePair rs) pairs
        where 
                pairs = List.concatMap (\ t -> map (head t, ) (tail t)) . filter (length >>> (> 1)) . tails $ u

middleElement :: [a] -> a
middleElement us = (us !!) $ (length us `div` 2)

part1 (rs, us) = sum . map (middleElement) . filter (validateUpdate rs) $ us

insertOrderly rs i is = insertOrderly' frontRules i is
        where
                frontRules = filter (((== i) . fst)) rs

insertOrderly' _  i [] = [i]
insertOrderly' rs i (i':is)
        | any (snd >>> (== i')) rs = i : i' : is
        | otherwise = i' : insertOrderly' rs i is

part2 (rs, us) = sum . map middleElement . Maybe.mapMaybe ((orderUpdate &&& id) >>> \ p -> if (fst p /= snd p) then Just $ fst p else Nothing) $ us
        where
                orderUpdate = foldr (insertOrderly rs) []

main = getContents >>= print . (part1 &&& part2) . parse