If I have compact fiber bunlde $E\to M$ with fiber $F$ and differential $j$-form $\alpha$ on $F$, from which I know that it extends to $E$. For example, $E$ is obtained from a group action on $F \times I$ under which $\alpha$ is invariant. Is the extension of $\alpha$ then unique?
The motivation for this question is:
I know from a Mayer-Vietoris argument, that $i^*:H^j(F;\mathbb R) \to H^j(E;\mathbb R)$ is injective. Does this allow me to make conclusions on the level of forms?
No, the extension is not unique in general. Here is a very simple case. Consider $F = \{0\}$ to be a singleton, and $E = \mathbb{R}$ (and $M = \mathbb{R}$, the bundle is the identity). You are asking if given a value $a \in \mathbb{R}$, there exists a unique function $f : E \to \mathbb{R}$ such that $f(0) = a$. That's clearly not the case.
Facts about the cohomology will generally only give you information about closed forms, and even then, only up to exact forms (that's pretty much the definition of cohomology).
Even with closed forms it doesn't work. Consider two different functions $f,g : \mathbb{R} \to \mathbb{R}$ whose derivatives vanish at zero, e.g. $f(x) = x^2$ and $g(x) = x^3$. Then $df$ and $dg$ are two closed $1$-forms on $E = \mathbb{R}$ which extend $0 \in \Omega^1(F) = \Omega^1(\{0\})$.