How is the sequence 1, 1.4, 1.41, 1.414 generated?

The following is an example of a method that you could use to figure this out. The method is called bisection. It is not the fastest, but it is clear how to extract digits from it because it has explicit error bounds.

You know $1^2<2$. You know $2^2>2$. So $1<\sqrt{2}<2$ so its first decimal digit is $1$.

You check $1.5^2>2$. You check $1.25^2<2$. You check $1.375^2<2$. You check $1.4375^2>2$. You check $1.40625^2<2$. Now you know $1.40625<\sqrt{2}<1.4375$ so you know the first two digits are $1.4$.

You continue: you check $1.421875^2>2$. You check $1.4140625^2<2$. You check $1.41796875^2>2$. So $1.4140625<\sqrt{2}<1.41796875$, so you know the first three decimal digits now.

You can keep going; at each time you know $\sqrt{2}$ is in between two numbers getting closer together, so as soon as those numbers have a new digit in common, you know that digit of $\sqrt{2}$. On average it takes $\log_2(10) \approx 3.3$ steps to get a new correct decimal digit.

At the cost of slightly more iterations, you can make calculations easier (if you're doing it by hand) by rounding the lower bound down and/or the upper bound up. For instance back in the second paragraph of iterations you could have said $1.4<\sqrt{2}<1.44$ and then continued, obtaining $1.41<\sqrt{2}<1.415$ at the end of the third paragraph.

In general, without more information one cannot produce a term in a sequence using just its previous terms.

One can describe where both of these particular sequences come from, however:

The first comes from truncating the decimal expansion of $\sqrt{2}$ at each successive decimal place. One can give an easy explicit formula for this:

$$a_n := 10^{-n} \lfloor 10^n \sqrt{2} \rfloor .$$

The second is similar, but instead of truncating, i.e., rounding down the nearest number whose decimal expansion has $\leq n$ digits, one rounds up:

$$a_n := 10^{-n} \lceil 10^n \sqrt{2} \rceil .$$

There are infinitely many more Cauchy sequences with limit $\sqrt{2}$. One uses the so-called Babylonian Method, which is itself a specialization of Newton's Method. In this case the terms are defined iteratively, by $$a_n := \frac{1}{2}\left(a_{n - 1} + \frac{2}{a_{n - 1}}\right) ,$$ where we can take $a_0$ to be any suitable value. Taking the convenient value $a_0 = 2$ gives the sequence $$2, \frac{3}{2}, \frac{17}{12}, \frac{577}{408}, \ldots ,$$ which converges relatively quickly.

Not the numbers in the question, but the integral

$$\int_0^1 \frac{x^m(1-x)^n}{\sqrt{1+x}}dx$$

produces rational approximations to $\sqrt{2}$ from above for $m$ odd and from below for $m$ even.

Some of the approximations for small $m,n$ include

$$\frac{1}{2}\int_0^1 \frac{1}{\sqrt{1+x}}dx =\sqrt{2} -1 $$ $$\frac{3}{2}\int_0^1 \frac{x}{\sqrt{1+x}}dx = 2 - \sqrt{2}$$ $$\frac{3}{8}\int_0^1 \frac{1-x}{\sqrt{1+x}}dx = \sqrt{2}-\frac{5}{4}$$ $$\frac{5}{8}\int_0^1 \frac{x(1-x)}{\sqrt{1+x}}dx=\frac{3}{2}-\sqrt{2}$$

$$\frac{21}{64}\int_0^1 \frac{x(1-x)^2}{\sqrt{1+x}}dx=\frac{23}{16}-\sqrt{2}$$

$$\frac{315}{832}\int_0^1 \frac{x^2(1-x)^2}{\sqrt{1+x}}dx = \sqrt{2}-\frac{73}{52}$$

Another way to obtain approximations converging to $\sqrt{2}$ comes from setting $x=1$ into the expansion for $\sqrt{1+x}$, which yields

$$\sqrt{2}=1+\frac{1}{2}-\frac{1}{8}+\frac{1}{16}-\frac{5}{128}+\frac{7}{256}...$$

This expansion and many more methods can be found in Pythagoras' Constant: $\sqrt{2}$ by Gourdon and Sebah. For instance, example 9, due to Euler,

$$\sqrt{2}=\frac{7}{5}\sum_{n=0}^\infty \frac{(2n)!}{(n!)^2 200^n}$$

produces exactly $1.4, 1.414, 1.41421$ when truncating to $1, 2$ and $3$ terms, and it provides about two correct decimal digits per term.

Finally, the Egyptian fraction described in the Wikipedia can be written in closed form using Paolo Lava's formula for sequence https://oeis.org/A082405.

$$\sqrt{2}=\frac{3}{2}-\sum_{k=0}^\infty \frac{2\sqrt{2}}{(17+12\sqrt{2})^{2^k}-(17-12\sqrt{2})^{2^k}}$$

The first approximations produced are $$\frac{3}{2},\frac{17}{12},\frac{577}{408},\frac{665857}{470832}, \frac{886731088897}{627013566048},\frac{1572584048032918633353217}{1111984844349868137938112},$$

so this may be a series equivalent form of Newton's iteration.