Part of your trouble may be related to the somewhat artificial splitting between $q$'s and $p$'s here. Let's take a $2$-dimensional configuration space with coordinates $(q^1,q^2)$. An arbitrary point in the cotangent bundle then has the four coordinates $x \equiv (q^1,q^2,p_1,p_2)$, so $x^1=q^1,x^2=q^2,x^3=p_1,x^4=p_2$.
In this language, the canonical $2$-form can be written
$$\omega= \mathrm dx^3\wedge \mathrm dx^1 + \mathrm dx^4 \wedge \mathrm dx^2 $$
$$\equiv \frac{1}{2}\big(\mathrm dx^3\otimes \mathrm dx^1 - \mathrm dx^1 \otimes \mathrm dx^3 + \mathrm dx^4 \otimes \mathrm dx^2 - \mathrm dx^2 \otimes \mathrm dx^4\big)$$
If we feed this $2$-form two arbitrary vectors $A = \sum_{n=1}^4 A^n \frac{\partial}{\partial x^n}$ and $B = \sum_{n=1}^4 B^n \frac{\partial}{\partial x^n}$, we obtain
$$\omega(A,B)= \frac{1}{2}\big(A^3B^1-A^1B^3 + A^4B^2 - A^2 B^4\big)$$
Recall that $\omega$ is non-degenerate $\iff \omega(A,B)=0$ for all $B$ implies that $A=0$. If the expression $\omega(A,B)$ written above is equal to zero for every possible vector $B$, what does that tell you about $A$?
In my attempts I used another vector field that I found out now was the reason I wasn't able to prove it, but would it be that I could prove it with this vector field?
$$X(f) =\partial_{p{\mu}}f \partial_{q^{\mu}} - \partial _{q^{\mu}}f \partial_{p{\mu}}$$
To better understand the notation
If $$f(q, p) =H(q, p) (hamiltonian) $$
$$X(H) =\frac{d} {dt} $$
I believe I understand your follow-up question. Let the phase space be given by $X$. Given a non-degenerate $(0,2)$-tensor $B:TX\times TX\rightarrow \mathbb R$, one can define a map $\flat: TX \rightarrow T^*X$ via $ A^\flat := B(A,\cdot)$, where $\cdot$ denotes an empty slot. In component form,
$$(A^\flat)_\mu = B_{\mu\nu} A^\nu$$
If we define $\tilde B$ to be the $(2,0)$-tensor whose components are the matrix inverse of the components of $B$ (by which I mean, $\tilde B^{\mu\alpha} B_{\alpha\nu} = \delta^\mu_\nu$) then $\tilde B$ defines the inverse map
$$\sharp : T^*X \rightarrow TX$$
$$(\alpha^\sharp)^\mu = \tilde B^{\mu\nu} \alpha_\nu$$
This association between the tangent vectors and covectors is called a musical isomorphism, and it is well-defined if and only if $B$ is non-degenerate (so that the matrix $B_{\mu\nu}$ is invertible).
With that out of the way, to each smooth function $f$ corresponds a so-called Hamiltonian vector field $X_f$ given by $X_f := (\mathrm df)^\sharp$, where the musical isomorphism is provided by $\omega$. In canonical coordinates, one has that
$$f \mapsto X_f = \sum_n\frac{\partial f}{\partial p_n} \frac{\partial}{\partial q^n} - \frac{\partial f}{\partial q^n} \frac{\partial}{\partial p_n}$$
The association $\mathrm df \leftrightarrow X_f$ is an isomorphism if and only if $\omega$ is non-degenerate. In explicit component form (in canonical coordinates), the association takes the form
$$\underbrace{\left(\frac{\partial f}{\partial q^1},\frac{\partial f}{\partial q^2},\frac{\partial f}{\partial p_1}, \frac{\partial f}{\partial p_2}\right)}_{\text{components of }\mathrm df}\leftrightarrow \underbrace{\left(\frac{\partial f}{\partial p_1},\frac{\partial f}{\partial p_2},-\frac{\partial f}{\partial q^1}, -\frac{\partial f}{\partial q^2}\right)}_{\text{components of }X_f}$$
It's obvious that this is an isomorphism - the components are just shuffled around, modulo a minus sign - and so the bilinear form $\omega$ from which it is derived must be non-degenerate.
Best Answer
The name seems appropriate if consider that it probably comes from the case when the manifold is the cotangent bundle of a manifold. Then a point on $T^*M$ is a pair $(x,\alpha)$, where $x$ is a point on $M$ and $\alpha$ a one form. The definition of the tautological one form is: the value of the form at a point $(x,\alpha)$ on a tangent vector is obtained by projecting the vector to a tangent vector on $M$ and evaluating (at $x$) the form $\alpha$. In other words for $v\in T(T^*M)$ the value is $\theta_{(x,\alpha)}(v):=\alpha_x(d\pi(v))$, where $\pi : T^*M \rightarrow M$ is the projection.
In a way it is pretty tautological.