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 $w_1\in\Sigma^*$.
A recursive definition of $f$ (for any alphabet $A$) should be something like this:
$$f(n) = \begin{cases} \{\lambda\} &\mbox{if } n = 0 \\
A &\mbox{if } n = 1\\
\{ava : a \in A, v \in f(n - 2) \} &\mbox{if } n > 1
\end{cases}$$
Best Answer
If you try generating palindromes using the given definition, you'll get a set that looks something like: $\{\lambda, 00, 11, 0000, 1001, 0110, 1111, 000000, 100001, 010010, 110011, 001100, 101101, 011110, 111111, \dots \}$. In fact, all of the strings in this list have an even length. The reason why should be pretty clear - you're starting with a 0-length string, and always adding 2 more bits. So you'll never reach an odd-length string like $101$.
What are the shortest odd-length palindromes? They're the single bits, $0$ and $1$. You definitely can't reach them from the basis and the given recursion rule, so you either have to change the basis or the recursion to include them in the set. Which one makes more sense?