I've been searching the internet and books for a definition but none of the books on differential geometry and manifolds that I have contain the term rank in the index.
While trying to find a definition I came across this (very bottom of page 394):
The differential form (2.7) is of maximal rank since by the criterion of nondegenracy there is a nonzero exterior power $\Omega^n = \pm k dp_1 \wedge \dots dp_n \wedge \dots $ which up to the factor $\pm k$ is just the $2n$-volume form on $T^\ast M$.
It seems such a basic notion that I'm bewildered that I can't seem to find a definition. I'm very confused. I suspect that there is something wrong with the terms I search for but I don't know what else to search for.
What is a differential form of maximal rank and what is the rank of a differential form?
I'm even confused about the most basic examples:
Consider for example differential one forms $\alpha, \beta, $. Then the differential form $\gamma = \alpha \wedge \beta$ has rank one because the minimal number of terms in the sum that constitutes $\gamma$ is just one term, namely, $\alpha \wedge \beta$. But the rank of $\gamma$ is two: it is a volume form in $\mathbb R^2$ and the maximal rank of any differential form in $\mathbb R^2$ is 2.
The notion of rank you are looking for admits an interpretation as the tensor rank for form as in the comment of @John_Ma, but I think this is rather confusing. The example your are looking at is the case of a two--form, which, in each point, defines a skew symmetric bilinear form on the tangent space. Hence you can look at the (point-wise) rank of this bilinear form. In particular, "is of maximal rank" in this case just means "is non-degenerate in each point", i.e. if $\omega (\xi,\eta)=0$ for all $\eta$, then $\xi=0$. This is only possible in the case of even dimension, and if the dimesnion equals $2n$, then non-degeneracy is equivalent to the fact that the $n$-fold wedge product of $\omega$ with itself is non-zero and hence a volume form.