Roberto thanks!
Shame on me, to post code without enough testing :(
Yet, thanks to your comments *I think* I have found the bugs you wrote about and now my code works, please see corrected version below.
Extra substring at the end was a result of using foldr with initial element of []. I fixed this with foldl and first chunk as its initial element.
Incomplete substitution in case of duplicate elements in the pattern was a bug in my 'takeOut' function that I have also fixed.
Stiil the problem that I have not yet designed solution for is when a substring to replace extends from the end of a string to the next string. In other words - first part of substring ends the first string and second part of substring starts the second string. My algorithm currently does not account for such a case.

On the side: The more I use Haskell - the more I like it ! It helps me think about the problem I solve much more clearly then when I use imperative language.

Corrected code:

-- replace all occurances of "123" with "58" in a string:
test = replStr "abc123def123gh123ikl" "123" "58"

{--
In a string replace all occurances of an 'old' substring with a 'new' substring
--}
replStr str old new = foldl ((\newSub before after -> before ++ newSub ++ after) new) firstChunk otherChunks
                      where chunks = splitStr str old
                            firstChunk = head chunks
                            otherChunks = tail chunks
{--
Split string into a list of chunks.
Chunks are substrings located in a string between 'sub' substrings
--}
splitStr str sub = mkChunkLst str sub []
    where
      -- mkChunkLst 'src string' 'substr-to-extract' 'list of chunks'
      -- makes list of chunks located between 'substr-to-extract' pieces in src string
      mkChunkLst [] _ chunkLst = chunkLst
      mkChunkLst str sub chunkLst = mkChunkLst after sub (chunkLst ++ [chunk])
          where
            (chunk, _, after) = takeOut str sub [] []

{--
Take out substring from a string.
String is divided into:
"before substr" ++ "match" ++ "after substr"
where 'match' is substring to split out 
--}

takeOut after [] before match = (before, match, after)
takeOut [] _ before match = (before, match, [])
takeOut (x:xs) (y:ys) before match
       | x == y = takeOut xs ys before (match ++ [x])
       | otherwise = takeOut xs (y:ys) (before ++ [x]) []






--
Dmitri O. Kondratiev
dokondr@gmail.com
http://www.geocities.com/dkondr