Interior product of differential forms

It's useful to think of differential forms as antisymmetric multilinear mappings, i.e a $p$-form eats $p$ arbitrary vector fields and gives you an ordinary function of $n$ variables (where $n$ is the dimension of your manifold).

The interior product is a mapping from a $p$ form $\omega$ to a $(p-1)$ form, since you've fixed one argument of the $p$ form to be a particular vector field, say $X$ and as a result, $\iota_X \omega$ can only act on $(p-1)$ vector fields now.

To get the explicit formula in local coordinates for the interior product of a form with a vector field, you can use the formula for two forms $$(\iota_X \eta)_i = X^j\eta_{ji}$$(summation convention) to get $\iota_X d\omega$:

$(i_X d\omega) = X^j (d\omega)_{ji} dx^{i}.$

If this is still confusing, consider working your example out directly by the definition of everything:

$d\omega = 3dx\wedge dy - 14zxdx\wedge dz$, which written in terms of tensors is given by

$d\omega = 3(dx\otimes dy - dy \otimes dx) - 14zx (dx\otimes dz - dz\otimes dx)$

So then contracting in the first argument with the vector field X gives you:

$\iota_X d\omega = 3(dx(X)\otimes dy - dy(X) \otimes dx) - 14zx (dx(X)\otimes dz - dz(X)\otimes dx)$

$\iota_X d\omega = 3[dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dy - dy(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) \otimes dx] - 14zx [dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dz - dz(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z})\otimes dx]$

Now since $dx$ is a linear functional (field) that are dual to $\partial_x$ (and so on for $y$ and $z$), we get that

$dx(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = y$

$dy(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = 2z$

$dz(y\frac{\partial}{\partial x}+2z\frac{\partial}{\partial y}+3xy\frac{\partial}{\partial z}) = 3xy$.

Using this, gives you

$\iota_X d\omega = [-3(2z)+ 14zx(3xy)]dx + (3y)dy - (14zxy)dz$

as your one-form.

Note that this is a laborious way and only serves to illustrate the definitions to you. In practice, it would be faster to get the answer by the formula given above.

Now you can take any arbitrary vector field $Y$, do the same contraction with this now one-form to get a function of $3$ variables.

$Y$ is arbitrary (in the sense that it could be any vector field of the correct dimension; if you have a 2-form built on $\mathbb R^3$, then the vector field should be $\mathbb R^3 \mapsto \mathbb R^3$).

So what you're discovering is that the interior product of a vector field and a 2-form field generates a 1-form field. Let $\sigma(Y) = i_X (d\omega)(Y)$. Then we have,

$$\sigma(Y) = d\omega(X, Y)$$

It should be clear looking at the left-hand side that this is indeed a 1-form.

What is the interior product of a one-form and a vector? Well, one-forms take vectors as arguments and return scalars. That's part and parcel to the idea of one-forms as linear functionals.