Haskell's type system is really expressive, so I suggest to think about the problem in terms of types. The advantage of this is that you can solve the problem 'top-down' and the whole program can be typechecked as you go, so you can catch all kinds of errors early on. The general approach is to incrementally divide the problem into smaller functions, each of which remaining undefined initially but with some plausible type.
What you want is a function (let's call it convert) which take a list of strings and generates a list of tuples, i.e.
convert :: [String] -> [(String, String)]
convert = undefined
It's clear that each string in the input list will need to be parsed into a 2-tuple of strings. However, it's possible that the parsing can fail - the sheer type String makes no guarantees that your input string is well formed. So your parse function maybe returns a tuple. We get:
parse :: String -> Maybe (String, String)
parse = undefined
We can immediately plug this into our convert function using mapMaybe:
convert :: [String] -> [(String, String)]
convert list = mapMaybe parse list
So far, so good - but parse is literally still undefined. Let's say that it should first verify that the input string is 'valid', and if it is - it splits it. So we'll need
valid :: String -> Bool
valid = undefined
split :: String -> (String, String)
split = undefined
Now we can define parse:
parse :: String -> Maybe (String, String)
parse s | valid s   = Just (split s)
        | otherwise = Nothing
What makes a string valid? Let's say it has to contain a = sign:
valid :: String -> Bool
valid s = '=' `elem` s
For splitting, we'll take all the characters up to the first = for the first tuple element, and the rest for the second. However, you probably want to trim leading/trailing whitespace as well, so we'll need another function. For now, let's make it a no-op
trim :: String -> String
trim = id
Using this, we can finally define
split :: String -> (String, String)
split s = (trim a, trim (tail b))
  where
    (a, b) = span (/= '=') s
Note that we can safely call tail here because we know that b is never empty  because there's always a separator (that's what valid verified). Type-wise, it would've been nice to express this guarantee using a "non-empty string" but that may be a bit overengineered. :-)
Now, there are a lot of solutions to the problem, this is just one example (and there are ways to shorten the code using eta reduction or existing libraries). The main point I'm trying to get across is that Haskell's type system allows you to approach the problem in a way which is directed by types, which means the compiler helps you fleshing out a solution from the very beginning.