Whatever you tried to draw didn't render at all for me. But anyway.
Let's say you're adding 24 + 26. If you're familiar with positional numerals -- that is, where the value of a digit depends on where it's written in the number -- you'll know that 24 is 2 10's + 4 1's, and 26 is 2 10's + 6 1's. Let's add them: how many 10's? 2 from 24 plus 2 from 26 give you 4 10's. How many 1's? 4 from 24 and 6 from 26 give you 10 1's. But 10 1's is the same as 1 10, so we add that to our total of 10's, giving us 5 10's, which we write as 50.
So what did we do? We added up the 10's, then we added up the 1's, then we combined them into one result. But when we combined them, we actually had to *change* the number of 10's we had because the number of 1's was too big for one digit. That's annoying to do, so we should probably add up the 1's first and then the 10's, since the answer to the sum of 1's could affect the number of 10's in the answer, but the sum of 10's can't affect the number of 1's.
Now we have an algorithm, a systematic way to add two numbers by adding their digits. First we add the 1's, and if there's a number of 10's left over, we add *that* to the sum of the 10's, which, if there's a number of 100's left over, we add that to the sum of the 100's, etc. But how do we remember the steps? Easiest way is to write the numbers vertically, lining up the 1's, the 10's, etc. This way, we can just add down the 1's column and write the answer, remembering to keep track of leftover 10's; then we add the 10's column (plus any leftover 10's) and write that answer, remembering to keep track of leftover 100's, and so on. You don't *have* to write the numbers that way, and you don't *have* to add the 1's first and work backwards. But if you don't, you'll have to erase your 10's if your 1's spill over. It's much easier to keep track of these spillover 10's by writing that number on top of the 10's column, so that you'll easily remember to add it.
Subtraction is exactly the same, except that in subtraction you might need to split a 10 up into 10 1's to make the subtraction work. If you have 24 and want to subtract 8, well, you can't take 8 1's from 4 1's, so you turn 24 into 1 10 and 14 1's and make your subtraction that way. You can easily represent this by sticking a little 1 next to the 4 and crossing out the 2 to write in a 1, which we call borrowing.
Multiplication is different. Here, it's easier to just do algebra. Let's say we're trying to multiply ABC by DEF (those are all digits). This is the same as doing ABC · (D·100 + E·10 + F·1). Multiplying ABC by the single digit F is the same as doing (A·100 + B·10 + C·1) · F = 100·A·F + 10·B·F + 1·C·F. So you multiply the 1's by F, and if there are any 10's left over, you add it to the total of the 10's times F, etc. Then you add them all up. It's all just the distributive property!
Division is completely different in form, but it works on a similar principle: let's say you're dividing 458 by 7. We'll go digit by digit. Note that 458 is 4 100's + 5 10's + 8 1's, right? How many 700's go into 458? 0, so the answer has a 0 in the 100's place. Subtract 0·700 = 0 from 458 and go to the 10's. How many 70's go into 458? 6, so the answer has a 6 in the 10's place. Subtract 6·70 = 420 from 458 to get 38 and go on. How many 7's go into 38? 5, so the answer has a 5 in the 1's place. Subtract 5·7 = 35 from 38 to get 3 and go on. Now, we could go to the 10ths place, 100ths place, etc., and so on forever, or we can declare that 3 is the remainder, since 7 doesn't go into it at all. The answer is therefore 65 remainder 3.
In all of these cases, it's very helpful in understanding the algorithms to keep track of what the digits actually represent. A 2 in a number could be 2, or 20, or 20000, or 0.00000002, depending on its position in the number. These algorithms are optimized in a way that will let you kind of ignore that information, which takes off some of the mental load and makes them really easy to actually execute on paper (other than long division, at least, since you can't do it digit by digit like you can with addition, subtraction, and multiplication). But when you think about these algorithms *with* the positional information, you can see them for what they are: just breaking down a number and organizing it to make things easier.
Hope that answers your question!