Reading numbers from input Haskell

As @phg points out, you are essentially constructing an infinite list, and actually evaluating it causes the loop error. A simple implementation to resolve this issue is to define a helper function which takes an additional parameter - a list to store all the inputs read in from the screen, like so:

readInteger :: IO ()
readInteger = readInteger' []
    where
        readInteger' x = do
        putStrLn "insert a number: "
        num<-getLine        
        if ((read num :: Int) == 0)  
        then print $ ordList x 
        else readInteger' $ (read num :: Int):x
        where ordList ::[Int]->[Int]
              ordList [] = []
              ordList xs = sort xs

Please note that the above is essentially just an implementation of @phg's answer, but with some changes to your original logic. Firstly, since 0 is a sentinel value, we shouldn't be appending that to our list. Second, we do not need to sort the list every single time we are adding a value to it. Sorting once at the time of printing/passing to another function is sufficient.

Demo

If you want to read an unspecified number of integers without prompting for user input and cut it off the moment you encounter 0, you would probably do well to use getContents, which will read everything from the standard input as a single string, lazily.

Then, it is a simple matter of parsing it to a list of numbers and doing what you want with it, like so:

readIntegers :: ()
readIntegers = do
   a <- getContents
   let b = ordList $ takeWhile (/= 0) $ map (\x -> read x :: Int) $ words a          
   mapM (putStrLn . show) b
 where ordList ::[Int]->[Int]
       ordList [] = []
       ordList xs = sort xs 

let list = ordList ((read num :: Int):list)

This is basically a recursive definition of a list of the form [x, x, ...] (like if you wrote an equation saying x = 1 + x). That is perfectly fine by itself, since Haskell is lazy; however, if you try to print list (aka "solve the equation"), it will fail, since it will try to print infinitely many numbers.

You probably have a misconception about the workings of the (:) operator. Haskell functions will never perform an assignment operation and concatenate num onto list by changing it, like in imperative languages. There are only pure functions.

If you want to accumulate all numbers, you should try to come up with a recursive definition of readIntegers, keeping its state (the list) in an additional parameter (there are also more sophisticated ways, hiding the state passing, but more complicated to use for a beginner).