palindromeCounter.js

function isPalindrome(str) {
  var reverse = str.split("").reverse().join("");
  return str === reverse;
}

function getCharCounts(chars) {
  var charCounts = {};
  for (var i = 0, charLen = chars.length; i < charLen; i += 1) {
    if (charCounts[chars[i]]) {
      charCounts[chars[i]] += 1;
    } else {
      charCounts[chars[i]] = 1;
    }
  }
  
  return charCounts;
}

function possiblePalindrome(str) {
  if (isPalindrome(str)) {
    return true;
  }
  
  var chars = str.split("");
  var charCounts = getCharCounts(chars);
  var oddsCount = 0;
  
  var charKeys = Object.keys(charCounts);
  for (var i = 0, len = charKeys.length; i < len; i += 1) {
    if (charCounts[charKeys[i]] % 2 === 1) {
      oddsCount += 1;
    }
    
    if (oddsCount > 1) {
      return false;
    }
  }
  
  return true;
}

function palindromeCounter(str) {
  str = str.toLowerCase();
  var words = str.split(" ");
  var counter = 0;
  
  for (var i = 0, len = words.length; i < len; i += 1) {
    if (possiblePalindrome(words[i])) {
      counter += 1;
    }
  }
  
  return counter;
}

console.log(palindromeCounter("a"));            // 1  ("a")
console.log(palindromeCounter("ab"));           // 0
console.log(palindromeCounter("Abc Def G"));    // 1  ("g")
console.log(palindromeCounter("abbAc C ddEf")); // 2  ("abcba", "c")
console.log(palindromeCounter("lmn opq rst"));  // 0
console.log(palindromeCounter("AAA bAbbB"));    // 2  ("aaa", "bbabb")

I think there may be an issue with your solution here? Although I'm not sure how the code could be fixed, maybe you could chime in?

console.log(palindromeCounter("abbAc C ddEf")); // 2 ("abcba", "c")
This line should return 3. As the three palindromes would be:
"abcba"
"bacab"
"c"