A real-valued function $f(x)$ is square integrable over $\Bbb{R}$, we write $f \in L^2(\Bbb{R})$, if and only if
$$ \int_{-\infty}^{+\infty} f(x)^2 dx < \infty \tag{1} $$
A necessary condition for the above integral to be finite is $$\lim_{\vert x \vert\to\infty} f(x) = 0$$
To convince yourself, think of the interpretation of an integral as the area under a curve: what do you think happens to the integral $(1)$ when $f(x)$ tends to a non-zero asymptotic limit, knowing that the integrand $f(x)^2$ is positive everywhere?
Let $C(k,T)$ denote the price (as of today) of a European call option, expiring at $T$ and struck at $K=e^k$. Since $$\lim_{k\to-\infty} C(k,T) = \lim_{k\to-\infty} \Bbb{E}^\Bbb{Q}_t \left[ (S_T - e^k)^+ B_T^{-1} \right] = B(0,T) F(0,T) \ne 0$$
we have that $C(k,T) \notin L^2$ which answers your first question.
Your second question is more technical. Basically, you are facing a situation where the risk-neutral pdf associated to your diffusion framework is not analytically tractable, so that you cannot evaluate the expression
$$ C(k,T) = \Bbb{E}^\Bbb{Q}_t \left[ (S_T - e^k)^+ B_T^{-1} \right] = \int_{-\infty}^{+\infty} (e^{s_T}-e^k)^+ B_T^{-1} \phi(T, s_T) ds_T $$
Still, your model being affine, you know that you can identify the characteristic function of $s_T=\ln(S_T)$ in closed-form, which happens to be the Fourier transform of the pdf $\phi(T,s_T)$:
$$\mathcal{G}_g(u) = \int_{-\infty}^{+\infty} e^{ius} \phi(T,s) ds$$
From there you would like to appeal to a result known as the Parseval relation, which would allow you to write
\begin{align}
C(k,T) &= \int_{-\infty}^{+\infty} \underbrace{(e^{s_T}-e^k)^+ B_T^{-1}}_{f(s_T)} \underbrace{\phi(T, s_T)}_{g(s_T)} ds_T \\
& = \langle f(s_T), g(s_T) \rangle = \frac{1}{2\pi}\langle \mathcal{F}_f(u), \mathcal{G}_g(u) \rangle \tag{Parseval} \\
&= \frac{1}{2\pi} \int_{-\infty}^{+\infty} \mathcal{F}_f(u) \mathcal{G}_g(u) du
\end{align}
which would allow you to exploit your knowledge of the characteristic function $\mathcal{G}_g(u)$: you just need to find the Fourier transform of the discounted payoff. This is exactly what @MJ73550 did in his answer.
The thing is that the Parseval relationship only holds for functions of $f$ and $g$ in $L^2$. Using a similar argument as above it is easy to see that $f \notin L^2$
