You could use a hash table for the inventory and check against and update currInv.

var curInv = [[21, "Bowling Ball"], [2, "Dirty Sock"], [2, "cat"], ],
    newInv = [[21, "Bowling Ball"], [2, "Dirty Sock"], [3, "rags"], [3, "mugs"]],
    inventory = Object.create(null);

curInv.forEach(function (a) {
    this[a[1]] = a;
}, inventory);

newInv.forEach(function (a) {
    if (!this[a[1]]) {
        this[a[1]] = [0, a[1]];
        curInv.push(this[a[1]]);
    }
    this[a[1]][0] += a[0];
}, inventory);

console.log(curInv);

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You are over-complicating things I think. Here's a working snippet.

function updateInventory(arr1, arr2) {
  for (var i = 0; i < arr2.length; i++) {
    var matchFound = false;
    for (var j = 0; j < arr1.length; j++) {
      if (arr1[j][1] === arr2[i][1]) {
        arr1[j][0] += arr2[i][0];
        matchFound = true;
      }
    }

    if (!matchFound) {
      arr1.push(arr2[i]);
    }
  }
  return arr1;
}


var curInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [2, "cat"],
];

var newInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [3, "rags"],
  [3, "mugs"]
];

console.log(updateInventory(curInv, newInv));

What it does

1. Loop through all of the elements in arr2.
2. Loop through all of the elements in arr1.
3. If a match is found, add the number of items in arr2 to the corresponding value in arr1.
4. If no match is found, add that element in arr2 to arr1.

You can merge the arrays Or combine the array so it will automatically do what you want .

var array3 = curInv.concat(newInv);

If you want to find the unique elements

// Merges both arrays and gets unique items var array3 = **

arrayUnique(array1.concat(array2));

function arrayUnique(array) {
   var a = array.concat(); 
   for(var i=0; i<a.length; ++i) { 
          for(var j=i+1; j<a.length; ++j) {
             if(a[i] === a[j]) a.splice(j--, 1); } 
     } 
   return a; 
}

**

Here is a concise functional programming style solution that uses a hash (a Map actually) for efficiency:

function updatedInventory(a, b) {
    return Array.from(
        b.reduce( (m, [v,k]) => m.set(k, (m.get(k) || 0) + v),
                   new Map(a.map ( ([v,k]) => [k,v] )) ), // swap pairs
        ([k,v]) => [v,k]) // swap back afterwards;
}

// Sample data
var curInv = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [2, "cat"],
];

var newInv = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [3, "rags"],
    [3, "mugs"]
];

// call the function
curInv = updatedInventory(curInv, newInv);
// Output the result
console.log(curInv);

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If your data would have pairs where the name is in the first element, and the number in the second, the code would have been much shorter. That is because Map objects are initialised with keys - value pairs, not value - key pairs:

function updatedInventory(a, b) {
    return [...b.reduce( (m, [v,k]) => m.set(k, (m.get(k) || 0) + v), new Map(a) )];
}

// Sample data
var curInv = [
    ["Bowling Ball", 21],
    ["Dirty Sock", 2],
    ["cat", 2],
];

var newInv = [
    ["Bowling Ball", 21],
    ["Dirty Sock", 2],
    ["rags", 2],
    ["mugs", 2]
];

// call the function
curInv = updatedInventory(curInv, newInv);
// Output the result
console.log(curInv);

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Because this function does not mutate the passed arguments, but returns the result, I have called it updatedInventory instead of updateInventory.

Issues with the code you tried with

1. your outer loop is on the first array, but that is not very practical, as you'll have a difficult time to identify the values that are missing in it. You should really loop over the second array first. Then you will find the elements that are missing in the first array.

2. There seems no good reason to start the inner loop at the index the outer one is currently at. So you'd better start at 0. Together with the first point, this would make your loops look like this:

   for (var j = 0; j < arr2.length; j++) {
     for (var i = 0; i < arr1.length; i++) {
   

3. When finding a match, there is no reason to look further, so you should then exit the inner loop:

       if (arr1[i][1] === arr2[j][1]) {
         arr1[i][0] += arr2[j][0];
         break;
       }
   

4. The second if you have, is really initiating a third loop, as indexOf performs an iteration. Furthermore, you are looking in arr1[i] which only has two elements: a value and a key. That was not your intension, I'm sure.

   Instead this code should be moved outside of the inner loop, and be corrected to look like this (again, taken into account your loops are swapped according to point 2):

     if (i >= arr1.length) { // this indicates no match was found
       arr1.push(arr2[j]);
     }
   

5. The third if was trying to align the length of the array, but it could well add an element that was already present in the other array! So that just has to go. In fact, the previous suggested correction would already make the arrays get aligned.

6. The break you have really killed it, and made you miss matches, and is the core reason (together with problem 4) why you got a duplicate. There is no reason to break here, as the first if might be true in one of the next iterations, so you should give that a chance still.