Why do we add a zero to dividend during long division?

The principle behind long division is repeated subtraction. If we want to divide a number $p$ by a number $q$, what we are trying to do is subtract as large a multiple of $q$ from $p$ as we can, until the difference is $0$ (if possible).

Let's take a nastier example for illustration. Suppose we wish to divide $4101$ by $12$. We want to subtract multiples of $12$ from $4101$. We can start by subtracting multiples of $12 \times 10^2 = 1200$, since this is a very simple, large multiple of $12$. We have \begin{align*} 4101 - \color{red}{3} \times 12 \times 10^2 &= 4101 - 3600 = 501 \ge 0 \\ 4101 - \color{red}{4} \times 12 \times 10^2 &= 4101 - 4800 = -699 < 0. \end{align*} So, we can only take $\color{red}{3}$ lots of $12 \times 10^2$ from $4101$ before we've taken too much. When we do, we are left with $501$. We know we cannot take even $1$ lot of $12 \times 10^2$, so we try one order of magnitude less: $12 \times 10^1$. We have \begin{align*} 501 - \color{green}{4} \times 12 \times 10^1 &= 501 - 480 = 21 \ge 0 \\ 501 - \color{green}{5} \times 12 \times 10^1 &= 501 - 600 = -99 < 0. \end{align*} We can only take $\color{green}{4}$ lots of $12 \times 10^1$ from $501$. Note that, in total we have $$4101 - \color{red}{3} \times 12 \times 10^2 - \color{green}{4} \times 12 \times 10^1 = 4101 - \color{red}{3}\color{green}{4}0 \times 12 = 21 \ge 0.$$ From $21$, we can take a single multiple of $12 \times 10^0 = 12$, as \begin{align*} 21 - \color{purple}{1} \times 12 \times 10^0 &= 21 - 12 = 9 \ge 0 \\ 21 - \color{purple}{2} \times 12 \times 10^0 &= 21 - 24 = -3 < 0. \end{align*} In total, $$4101 - \color{red}{3}\color{green}{4}\color{purple}{1} \times 12 = 9 \ge 0.$$

From the remaining $9$, $12$ cannot be subtracted any positive integer number of times without becoming negative. But, this doesn't mean we have to stop. Maybe no more whole amounts of $12$ will go into $9$, but we can try fractional parts of $12$. Again, all we do is go down an order of magnitude: \begin{align*} 9 - \color{orange}{7} \times 12 \times 10^{-1} &= 9 - 8.4 = 0.6 \ge 0 \\ 9 - \color{orange}{8} \times 12 \times 10^{-1} &= 9 - 9.6 = -0.6 < 0. \end{align*} So, $$4101 - \color{red}{3}\color{green}{4}\color{purple}{1}.\color{orange}{7} \times 12 = 0.6 \ge 0.$$ Once more, \begin{align*} 0.6 - \color{blue}{5} \times 12 \times 10^{-2} &= 0.6 - 0.6 = 0 \ge 0 \\ 0.6 - \color{blue}{6} \times 12 \times 10^{-2} &= 0.6 - 0.72 = -0.12 < 0. \end{align*}

That is, $$4101 - \color{red}{3}\color{green}{4}\color{purple}{1}.\color{orange}{7}\color{blue}{5} \times 12 = 0 \implies \frac{4101}{12} = \color{red}{3}\color{green}{4}\color{purple}{1}.\color{orange}{7}\color{blue}{5}.$$

I'm hoping this answer conveys a better understanding of the logic behind here. The decimal point in question is being introduced around where we subtract two numbers with a different number of decimal places. For example, when compute $$9 - 8.4 = 0.6 \ge 0.$$

Whether you write the decimal point and zero from the beginning, or after you realize you need them, is immaterial. They're always "there."

Or, just to be cautious, you could have written $3.0000000000$, which was more zeroes than you "needed" to solve this problem.

Decimals have infinitely many digits in both directions; when we only write finitely many of them, the intention is that every unwritten place has a zero.

When we do long division in the way you indicate, we have to actually use some of the places we had previously left unwritten. So we write in the zeroes.