This definition of concatenation is really just a more formal way of saying that in order to concatenate a string $w_1$ with a string $w_2$, we first append to $w_1$ the first character of $w_2$, then append to the result the second character of $w_2$, and continue in that fashion until we’ve exhausted $w_2$. For instance, it says that
$$\begin{align*}
abc\cdot def&=(abc\cdot de)f\\
&=\big((abc\cdot d)e\big)f\\
&=\big((abcd\cdot\lambda)e\big)f\\
&=\big((abcd)e\big)f\\
&=(abcde)f\\
&=abcdef\;.
\end{align*}$$
Note that it’s assumed that we know what it means to append a character to the end of a string; that’s what I did in the last two steps of the calculation.
I’ll prove by induction on the length of $w_2$ that the recursive definition really does tell you how to concatenate a string $w_1\in\Sigma^*$ with any string $w_2\in\Sigma^*$.
The basis step tells you how to concatenate a string $w_1$ with the string of length $0$, i.e., with the empty string: you just get $w_1$ back. Now suppose that you know how to concatenate $w_1$ with any string of length $n$; I claim that the induction step of the definition tells you how to concatenate $w_1$ with any string of length $n+1$. To see this, let $u$ be any string of length $n+1$. Then we can decompose $u$ into a string $w_2$ of length $n$ and a single character $x\in\Sigma$, so that $u=w_2x$. Then $$w_1\cdot w_2=w_1\cdot(w_2x)\;,$$ and the inductive step says that we can rewrite this as $$w_1\cdot w_2=w_1\cdot(w_2x)=(w_1\cdot w_2)x\;.$$ Since by hypothesis we already know how to concatenate $w_1$ with all strings of length $n$, we know how to form $w_1\cdot w_2$. It’s assumed from the beginning that we know what it means to append a single character to a string, so we also know what $(w_1\cdot w_2)x$ is, and therefore we know what $w_1\cdot u$ is. Finally, $u$ was an arbitrary string of length $n+1$, so we see that we now know how to concatenate $w_1$ with any string in $\Sigma^*$ of length $n+1$.
By induction, then, for each $n\in\Bbb N$ and each string $w_2\in\Sigma^*$ of length $n$ we know how to form $w_1\cdot w_2$. Finally, by definition every string in $\Sigma^*$ has length $n$ for some $n\in\Bbb N$, so we know how to form $w_1\cdot w_2$ for each $w_1\in\Sigma^*$.
Best Answer
The base case is the empty string, which is a member of S.
Now everything that has 2 zeros before an existing member of S is also a member of S.
Everything that has 1 one after an existing member of S is also a member of S.
This should give you every member of S.
(If you don't want to use the empty string as a base case because it's not a finite bit-string, you could also use two separate base cases, 00 and 1, and still apply the previous two rules.)