Let $X_t := t \cdot B_t$. By Itô's formula (applied to $f(t,x) := t^2 \cdot x^2$),
$$X_t^2-X_0^2 = 2\int_0^t s^2 B_s \, dB_s + \int_0^t (s^2 + 2s B_s^2) \, ds. \tag{1}$$
On the other hand,
$$X_s dX_s = s^2 B_s \, dB_s + s B_s^2 \, ds. \tag{2}$$
If we combine both equalities, we get
$$\langle X \rangle_t := X_t^2 - 2\int_0^t X_s \, dX_s = \int_0^t s^2 \, ds = \frac{t^3}{3}.$$
Let $X,Y$ be semimartingales and $\xi$ be an $X$-integrable process. Then,
$$\left[\int\xi\,dX,Y\right] = \int\xi\,d[X,Y]$$
and
$$\left[\int\xi\,dX\right]=\int\xi^2\,d[X]$$
where $[\cdot, \cdot ]$ denotes the quadratic covariation of two processes and $[\cdot ]$ denotes the quadratic covariation of a process with itself.
The proof of such statements is much easier if you use the following definition of $[\cdot]$:
Given a stochastic partition ${\mathcal P} = \left\{0=\tau_0\le\tau_1\le\tau_2\le\cdots\uparrow\infty\right\}$, we define
$$[X]^{{\mathcal P}}(t) \equiv \sum_{n=1}^\infty\left(X(\tau_n\wedge t)-X(\tau_{n-1}\wedge t)\right)^2$$
Taking a sequence of such partitions such that the mesh of the partitions is going to 0 in probability as $n$ increases, we define $[X]$ as the following limit
$$ [X]^{{\mathcal P}_n} \xrightarrow{ucp}[X]$$
The "ucp" convergence is the uniform convergence on compacts in probability.
Best Answer
Proof: For fixed $t>0$ choose a partition $\Pi = \{0=t_0 < \ldots <t_n=t\}$ of the interval $[0,t]$. Because of our assumptions on $f$, we know that $$s_j := f(t_j)$$ defines a partition $\Pi' := \{0=s_0 < \ldots < s_n = f(t)\}$ of the interval $[0,f(t)]$. Moreover, the continuity of $f$ implies that the mesh size $|\Pi'|$ of the partition $\Pi'$ tends to zero if $|\Pi|$ tends to zero. Therefore we find from the definition of the quadratic variation:
$$\begin{align*} \langle N \rangle_t = \lim_{|\Pi| \to 0} \sum_{j=1}^n (N_{t_j}-N_{t_{j-1}})^2 &= \lim_{|\Pi| \to 0} \sum_{j=1}^n (M(f(t_j))-M(f(t_{j-1})))^2 \\ &= \lim_{|\Pi'| \to 0} \sum_{j=1}^n (M(s_j)-M(s_{j-1}))^2 \\ &= \langle M \rangle_{f(t)}. \end{align*}$$
Proof: Denote by $L$ the Lipschitz constant of $g$, and let $\Pi = \{0=t_0 < \ldots < t_n\}$ be a partition of an interval $[0,t]$ for $t>0$. In order to compute the quadratic variation of $(Y_t)_{t \geq 0}$, we have to consider expressions of the form $$\sum_{j=1}^n (Y_{t_j}-Y_{t_{j-1}})^2 = \sum_{j=1}^n (g(t_j) X_{t_j}-g(t_{j-1}) X_{t_{j-1}})^2 = I_1+I_2+I_3$$ where $$\begin{align*} I_1 &:= \sum_{j=1}^n (g(t_j)-g(t_{j-1}))^2 X_{t_j}^2 \\ I_2 &:= -2\sum_{j=1}^n g(t_{j-1}) (g(t_j)-g(t_{j-1})) (X_{t_j}-X_{t_{j-1}}) \\ I_3 &:= \sum_{j=1}^n g(t_{j-1})^2 (X_{t_j}-X_{t_{j-1}})^2. \end{align*}$$ We estimate the terms separately. By the Lipschitz continuity of $g$ and the boundedness of $[0,t] \ni s \mapsto X_s(\omega)$ for fixed $\omega \in \Omega$, we have $$\begin{align*} |I_1| \leq L^2 \sup_{s \leq t} |X_s|^2 \sum_{j=1}^n \underbrace{(t_j-t_{j-1})^2}_{\leq |\Pi| (t_j-t_{j-1})} &\leq L^2 |\Pi| \sup_{s \leq t} |X_s|^2 \underbrace{\sum_{j=1}^n (t_j-t_{j-1})}_{t} \\ &\xrightarrow[\text{a.s.}]{|\Pi| \to 0} 0. \end{align*}$$Similarly, the Lipschitz continuity of $g$ and the uniform continuity of $s \mapsto X_s(\omega)$ on compact sets entails $$\begin{align*} |I_2| &\leq 2L \sup_{s \leq t} |g(s)| \sup_{\substack{u,v \in [0,t] \\ |u-v| \leq |\Pi|}} |X_u-X_v| \underbrace{\sum_{j=1}^n (t_j-t_{j-1})}_{t} \xrightarrow[\text{a.s.}]{|\Pi| \to 0} 0. \end{align*}$$ For the last term we note that $$\lim_{|\Pi| \to 0} \sum_{j=1}^n g(t_{j-1})^2 (X_{t_j}-X_{t_{j-1}})^2 = \lim_{|\Pi| \to 0} \sum_{j=1}^n g(t_{j-1})^2 (\langle X \rangle_{t_j}-\langle X \rangle_{t_{j-1}}) = \int_0^t g(s)^2 \, d\langle X \rangle_s; $$ this follows from the definition of the quadratic variation and basic properties of Riemann-Stieltjes integrals.
Using these two statements it shouldn't be too difficult to calculate the quadratic variation of $B$; if you, however, encounter any problems you are welcome to leave a comment.