For $k < 2$ and $k > \dim V - 2$ it's always true that a $k$-form $\omega$ both is decomposable and satisfies $\omega \wedge \omega = 0$.
For $k = 2$ (and provided the field underlying $V$ does not have characteristic $2$), the condition $\omega \wedge \omega = 0$ is both necessary and sufficient for decomposability. (Proving this is a nice exercise.)
The converse is not true in general, however. If $k$ is odd, then all $k$-forms $\omega$ satisfy $\omega \wedge \omega = 0$, but not all odd-degree multivectors (or, dually, forms) are decomposable:
Example If $\dim V \geq 5$, pick a basis $(E_a)$ and denote the dual basis by $(e_a)$. Then, the $3$-form
$$\psi := (e^1 \wedge e^2 + e^3 \wedge e^4) \wedge e^5$$
satisfies $\psi \wedge \psi = 0$ but we can show that it is indecomposable: Contracting a vector into a decomposable form yields a decomposable form. On the other hand, $$\iota_{E^5} \psi = e^1 \wedge e^2 + e^3 \wedge e^4 ,$$ and computing gives $(\iota_{E^5} \psi) \wedge (\iota_{E^5} \psi) \neq 0$, so by the criterion for $k = 2$, $\iota_{E^5} \psi$ is indecomposable.
For an algorithm that checks decomposability of a general $k$-form, see this old question.
Remark For $2 \leq k \leq \dim V - 2$, most $k$-forms are not decomposable, and we can quantify this assertion: For $0 \leq k \leq \dim V$, any $k$-vector $E_{a_1} \wedge \cdots \wedge E_{a_k}$ determines a $k$-plane in $V$, namely, $\langle E_{a_1}, \cdots, E_{a_k} \rangle$, and any $k$-plane in $V$ determines an underlying form up to an overall nonzero multiplicative constant. So, we may regard the space $D_k(V)$ of (nonzero) decomposable $k$-forms as a (punctured) line bundle over the space of all $k$-planes in $V$; this latter space is called the Grassmannian (manifold), $Gr(k, V)$, and it has dimension $k (\dim V - k)$, so $D_k(V)$ is a smooth manifold of dimension $k (\dim V - k) + 1$. On the other hand, the space of all $k$-forms has dimension $n \choose k$, and for $2 \leq k \leq \dim V - 2$,
$$\dim D_k(V) = k (\dim V - k) + 1 < {\dim V \choose k}$$ (but note that equality holds for $k = 1, \dim V - 1$).
Alternatively, the Plücker embedding realizes $Gr(k, V)$ as a projective variety in $\Bbb P(\Lambda^k V)$, and when $2 \leq k \leq \dim V - 2$ it is a proper subvariety, so its complement is nonempty and Zariski-open (and hence, when the underlying field is $\Bbb R$ or $\Bbb C$, dense with respect to the usual topology).
I think you may have forgotten what the symbol $dx$ means. Here $x$ is one of the coordinate functions in a chart at $p$, so it is a smooth function from some neighborhood of $p$ to $\mathbb{R}$. Then $dx$ is by definition the total derivative of $x$: that is, it is the functional on the tangent space at $p$ which takes a tangent vector $v$ at $p$ and outputs the directional derivative of the coordinate function $x$ with respect to $v$. Explicitly, if your local coordinates are $(x_1,\dots,x_n)$ with $x=x_i$ for some $i$ and you think of $v$ as a vector $(a_1,\dots,a_n)\in\mathbb{R}^n$ using these local coordinates, then $dx(v)$ will just be $a_i$. The functional $f(p)dx$ is then just this functional $dx$ multiplied by the scalar $f(p)\in\mathbb{R}$.
Best Answer
It is not true that any form whose exterior product with itself is zero must be decomposable. In fact, $\omega\wedge\omega = 0$ for all odd-degree forms $\omega$. This is because of the alternating nature of the exterior product: for any $r$-form $\omega$, we have $\omega\wedge\omega = (-1)^r\omega\wedge\omega$, which implies both sides are zero if $r$ is odd.
The claim is not true for even $r\neq 2$, either. For example, if $m \geq 7$, then
$\eta = v_1\wedge v_2 \wedge v_3\wedge v_4 + v_1\wedge v_5 \wedge v_6\wedge v_7$,
where $v_1,\dots ,v_7$ form a linearly independent set, is not decomposable, and yet $\eta \wedge\eta = 0$. This can be extended to other even forms. Also note in the above example with $m=7$, all 4-forms satisfy $\omega\wedge\omega = 0$, as there are no 8-forms.
In other words, the reverse of what you're saying is indeed not true for anything other than a $2$-form.