Let $\pi: E \rightarrow M$ be a vector bundle of rank $n$ and let $j: M \rightarrow E$ be its zero section.
Let $\beta$ be a closed differential form. I was convinced that $j^* \beta$ is the zero form, until I found the theorem which says that if $\beta$ is closed differential form in $\Omega(E)$, then $${(2 \pi)}^{-n/2}\int_E U \wedge \beta = \int_M j^*\beta$$ Where $U$ is a Thom form for the vector bundle $E$.
After reading this theorem I realize that $j^* \beta$ may not be the zero form. Please someone explains to me what is the form $j^* \beta$ ?
As I mentioned in the comments, $j^*\beta$ is not necessarily the zero form. The point is that for $X \in T_mM$, we need not have $j_*(X) = 0 \in T_{j(m)}E = T_{0_m}E$.
An illustrative example is the trivial line bundle $\pi: \mathbb{R}^2 \to \mathbb{R}$, $\pi(x, y) = x$ where now $j : \mathbb{R} \to \mathbb{R}^2$ is given by $j(x) = (x, 0)$. If $\beta$ is a $1$-form, then $\beta = fdx + gdy$ for some smooth functions $f, g : \mathbb{R}^2 \to \mathbb{R}$, so
$$j^*\beta = (f\circ j)d(x\circ j) + (g\circ j)d(y\circ j) = (f\circ j)dx.$$
Note that $(f\circ j)(x) = f(j(x)) = f(x, 0)$ which can be an arbitrary function of $x$, so every one-form on $\mathbb{R}$ is of the form $j^*\beta$ for some one-form $\beta$ on $\mathbb{R}^2$.