You can use the filter method for that:

const removeFromArray = function (arr, ...theArgs) {
  return arr.filter( val => !theArgs.includes(val) )
};

const list = [1,2,3];
const newList = removeFromArray(list, 2,3);

console.log(newList);

And a more terse version:

const removeFromArray = (arr, ...args)=> arr.filter( val => !args.includes(val) )

Tip: try to avoid mutating the original array and work on or return a copy during these operations.

The issue is with your indexing, you are finding the element using the index of arr, and deleting in the array, which is probably causing issue with indexing in loop. Modify your code as follows

const removeFromArray = function (arr, ...theArgs) {
  for (let i = 0; i < theArgs.length; i++) {
    if (arr.includes(theArgs[i])) {
      arr.splice(arr.indexOf(theArgs[i]), 1);
    }
  }
  return arr;
};

The above fixes the code your way, but a better way of doing it would be using filter.

const removeFromArray = function (arr, ...theArgs) {
  return arr.filter(ele => !theArgs.includes(ele))
}

I am writing it this way to purely maintain your function.

The problem is because you remove item from the array while being looping from that array.

Every time your for loop iterate the array, it will get a new array

e.g. (1,2,3,4,5 => 2,3,4,5), but the i value just keeping increasing by 1.

const removeFromArray = function (arr, ...theArgs) {
  for (let i = 0; i < arr.length; i++) {
  console.log(`arr:${arr}`,`item${arr[i]}`,`num${i}`)
  console.log(arr[i])
    if (theArgs.includes(arr[i])) {
      arr.splice(i, 1);
    }
  }
  return arr;
};
const testarray = [1,2,3,4,5]
console.log(removeFromArray(testarray,1,2,3))