Algorithms for addition work by taking advantage of the fact that a string of digits can be expanded (e.g. $123 = 1\times 10^2 + 2\times 10 + 3$) and by means of this expansion together with the algebraic properties of addition it can be "creatively reinterpreted" (e.g. $3$ is less than $4$? Well we can reinterpret the $2$ as $2$ $10$s etc...).
But how do traditional algorithms for square roots that they used to teach schoolchildren in the 19th century work?
I'm not exactly sure which algorithms you are referring to, but they could be based off of the Taylor Expansion. For example, if $a$ is small compared to $x$, you know that $$\sqrt{x+a} \approx \sqrt{x} + \frac{a}{2 \sqrt{x}}.$$ So then you could approximate $$\sqrt{103} \approx 10 + \frac{3}{20}.$$ You can get a higher degree of accuracy by taking more terms from the Taylor Expansion.
To see how the digit-by-digit method works, it might work best by stepping through an example. Let's set $N = 57$, and our aim is to find the sequence of digits $d_0, d_1, \ldots$ such that $\sqrt{N} = \sum_{n = 0}^\infty d_n 10^{-n}$. Notice that for every $n$, we want $d_0.d_1 \ldots d_n \leq \sqrt{N} < d_0.d_1 \ldots (d_n + 1) = d_0.d_1 \ldots d_n + 10^{-n}$.
So to find $d_0$, we're just looking for a number $a$ such that $a^2 \leq N < (a+1)^2$, and it's pretty clear that $a = 7$ works because $7^2 = 49 \leq 57 < 64 = 8^2$.
Then to find $d_1$, we now have $(7 + \frac{d_1}{10})^2 \leq 57 < (7 + \frac{d_1}{10} + \frac{1}{10})^2$. If we subtract $7^2$ from all of that, we get:
$$\begin{eqnarray} (7 + \frac{d_1}{10})^2 - 7^2 & \leq & 57 - 7^2 & < & (7 + \frac{d_1 + 1}{10})^2 - 7^2 \\ 49 - 2 \times 7 \times \frac{d_1}{10} + \frac{d_1^2}{10^2} - 49 & \leq & 57 - 49 & < & 49 - 2 \times 7 \times \frac{d_1 + 1}{10} + \frac{(d_1 + 1)^2}{10^2} - 49 \\ \frac{d_1}{10}\left(14 + \frac{d_1}{10}\right) & \leq & 8 & < & \frac{d_1 + 1}{10}\left(14 + \frac{d_1 + 1}{10}\right) \end{eqnarray}$$
So we're now looking for the biggest value for $d_1$ that keeps that left-hand inequality true, since the right-hand inequality is the same thing but using $d_1 + 1$ instead. With some trial and error we'll find that $d_1 = 5$ is right.
We can keep doing the same thing - once we have our $n$th approximation $a_n = d_0.d_1\ldots d_n$, we want to find the next one using the bound $(a_n + d_{n+1} \times 10^{-n-1})^2 \leq N < (a_n + (d_{n+1} + 1) \times 10^{-n-1})^2$, and that then becomes:
$$\begin{eqnarray} (a_n + d_{n+1} \times 10^{-n-1})^2 & \leq & N & < & (a_n + (d_{n+1} + 1) \times 10^{-n-1})^2 \\ \left(a_n^2 + \frac{d_{n+1}}{10^{n+1}}\left(2 a_n + \frac{d_{n+1}}{10^{n+1}} \right) \right) - a_n^2 & \leq & N - a_n^2 & < & \left(a_n^2 + \frac{d_{n+1} + 1}{10^{n+1}}\left(2 a_n + \frac{d_{n+1} + 1}{10^{n+1}} \right) \right) - a_n^2 \\ \frac{d_{n+1}}{10^{n+1}}\left(2 a_n + \frac{d_{n+1}}{10^{n+1}} \right) & \leq & N - a_n^2 & < & \frac{d_{n+1} + 1}{10^{n+1}}\left(2 a_n + \frac{d_{n+1} + 1}{10^{n+1}} \right) \\ \end{eqnarray}$$
So in other words we look at the gap between $a_n^2$ and $N$, and look for the value of $d_{n+1}$ such that the term on the left comes closest to matching that residual gap without going over. And what you'll find (but I won't prove rigorously here) is that the restriction of $d_n$ to single digits means that you will be able to work to within $(2n)$ decimal places, which is why in the "long division"-like method you pull down two digits at a time.
Sqrt(x) = x / sqrt(x).
You can calculate the square root with pen and paper using a method quite similar to long division, except that with every additional digit the divisor changes.
Two ways I can think of are the Newton's method (improved) and iterative methods
Suppose we want to solve $\sqrt{a}$
We try to solve the equation
$$ f(x) = \frac{1}{x^2} - a $$ with the iteration step: $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{\frac{1}{x_n^2}-a}{-2\frac{1}{x_n^3}}= \\ =x_n-\frac{x_n-x^3_na}{-2}=\frac{3}{2}x_n-\frac{x_n^3a}{2} $$
We check whether the number $x$ is a square root of $a$, by simply multiplying it with itself and with error $|a-x^2|$. The next part after the initial guess, we need to know if we want to increase the value of x or decrease it (which way do we want to take the guesses), and we use newton's method to do that: $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. $$ where $f(x)=x^2-a$ giving us: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n^2-a}{2x_n}= \\ =\frac{1}{2}x_n+\frac{a}{2x_n} $$
