Ito's formula, and the relationship between dt and dB(t) reading Bernt Oksendal Stochastic Differential Equations. 
I have just seen Ito's Formula, after this the author then says where $dX(t)^{2}$ is calculated using
 $$dt\cdot dt=dt \cdot dB(t) =dB(t)\cdot dt =0$$
 and
 $$dB(t) \cdot dB(t) = dt$$
On page 45 at the top. Where do these relations come from, what is the intuition behind it. 
 A: For suitable $\mu$ and $\sigma$, remember that
$$
dX_t = \mu(t, X_t) dt + \sigma(t, X_t) dB_t, \qquad X_0 = x,
$$
is shorthand for
$$
X_t = x + \int_0^t \mu(s, X_s) ds + \int_0^t \sigma(s, X_s) dB_s.
$$
Itô's formula for a suitable $f$ tells us that
$$
f(t, X_t) 
= f(t, x) + \int_0^t f_t(s, X_s) dt + \int_0^t f_x(s, X_s) dX_t 
+ \frac 1 2 \int_0^t f_{xx}(x, X_s) d \langle X \rangle_t,
$$
where $\langle X \rangle = (\langle X \rangle_t)_{t \geq 0}$ is the quadratic variation process of $X$ and is equal to
$$
\langle X \rangle_t 
= \left \langle \int_0^\cdot \sigma(s, X_s) d W_s \right \rangle_t
= \int_0^t \sigma^2(s, X_s) d \langle W \rangle_s
= \int_0^t \sigma^2(s, X_s) ds.
$$
The multiplication rules in Øksendal are stated in the way you wrote, so that the version of Itô's formula in that book yields the correct result. In essence, this is because the quadratic variation for a Brownain motion is $\langle B \rangle_t = t$. It is a good exercise to verify this for yourself. I hope this answers your question.
A: $dt\cdot dt=0$ is shorthand notation for the fact that the quadratic variation of the identity function with itself is $0$.
$dt\cdot dB(t)=0$ is shorthand notation for the fact that the cross variation of Brownian motion with the identity function is $0$.
$dB(t)\cdot dB(t)=dt$ is shorthand notation for the fact that the quadratic variation of Brownian motion with itself over a time interval of length $t$ is equal to $t$.
