I know that the Brownian motion is in $\lambda^2_\text{loc}$(namely the integrand $\phi$ is progressive and for every $t$, $\int_0^t \phi_s^2\,ds< \infty$), however, for every $t > 0$ it is in $\lambda^2(t)$ (the integrand is progressive and $\operatorname E\left[\int_0^t\phi_s^2\,ds \right] < \infty$) and finally that the B.M. it is not in $\lambda^2$ $\big(E\left[\int_0^\infty\phi_s^2\,ds \right] < \infty\big)$.
Therefore if I have a stochastic integral as the following: $$\int_0^tB_s\,dB_s$$ in this case is $B_s$ in $\lambda^2(t)$ because the interval of the integral is only on $[0,t]$, and so is sufficient only checking that $E\left[\int_0^t\phi_s^2\,ds \right] < \infty$, and not also that $E\left[\int_0^\infty\phi_s^2\,ds \right] < \infty$, right?. However, if I have $$\int_0^\infty B_s \, dB_s$$ in this case instead $B_s$ (the integrand) is in $\lambda^2_\text{loc}$ because I have the integral up to $\infty$. Is it right this distinction?
I would really appreaciate a feedback
To be honest, I do not fully understand what you are asking...but nevertheless, I will try to clear things up a bit.
Following your notation, I write
Then $$\lambda^2 \subseteq \bigcap_{t \geq 0} \lambda^2(t) \subseteq \lambda^2_{\text{loc}}. \tag{1}$$
Example 1: Consider $$\int_0^{\infty} B_s \, dB_s,$$ i.e. $\phi(s)=B_s 1_{[0,\infty)}(s)$. Then $\phi \notin \lambda^2$ since $\mathbb{E}\int_0^{\infty} B_s^2 \,d s = \int_0^{\infty} s \, ds=\infty$. However, $\phi \in \lambda^2(t)$ for each $t \geq 0$, and so, by $(1)$, $\phi \in \lambda^2_{\text{loc}}$.
Example 2: Consider for fixed $T>0$ $$\int_0^T B_s \, dB_s,$$ i.e. $\phi(s)=B_s 1_{[0,T]}(s)$. Then $\phi \in \lambda^2$ since $$\mathbb{E}\int_0^{\infty} \phi(s)^2 \, ds = \mathbb{E}\int_0^T B_s^2 \, ds = T < \infty.$$ In particular, $\phi \in \bigcap_{t \geq 0} \lambda^2(t)$ and $\phi \in \lambda^2_{\text{loc}}$.