(Below, by "vector space" I always mean "finite-dimensional real vector space.")
For $X$ any topological space there is a notion of a vector bundle over $X$ which formalizes the idea of a family of vector spaces parameterized by $X$. Vector bundles can be organized into a category similar to the category of vector spaces (which one recovers by taking $X$ to be a point); in particular one can define direct sums, duals, and tensor products of vector bundles.
An important notion here is that of a section of a vector bundle, which is roughly speaking a continuous choice of vector in each vector space in the family.
If $X$ is a smooth manifold (in particular if $X$ is an open subset of $\mathbb{R}^n$), then we can define a distinguished (smooth) bundle on $X$, the tangent bundle $T(X)$, coming from the tangent spaces at each point. The (smooth) sections of the tangent bundle are precisely vector fields on $X$. There is a dual bundle $T^{\ast}(X)$, the cotangent bundle, whose (smooth) sections are precisely differential forms on $X$.
So the tangent and cotangent bundles are in fact dual bundles, which means they have a dual pairing, and taking sections gives a dual pairing between vector fields and differential forms.
If in addition $X$ is compact, then we can make use of the Serre-Swan theorem, which identifies vector bundles over $X$ with finitely-generated projective modules over the ring $C^{\infty}(X)$ of smooth functions $X \to \mathbb{R}$. In this context I believe it's still true that the module of vector fields and the module of differential forms are dual, but I haven't worked out the details.
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.
Best Answer
Interior product is defined like this: if $\omega$ is a $k$-form and $X$ is a vector field, then $\iota_X \omega$ is a $(k-1)$-form defined by (remember, $\iota_X \omega$ "eats" $k-1$ vectors and returns a function): $$ [\iota_X \omega](V_1, V_2, \dots, V_{k-1}) := \omega(X,V_1, V_2, \dots, V_{k-1}) ,$$ i.e., we just put $X$ in the first argument of $\omega$, leaving $k-1$ remaining slots.
Here's an example: let's say $\omega$ is the two-form $\omega = dx \wedge dy$. Let's first just work out what $\omega$ is on some pairs of vector fields (you should essentially take all of the following to be part of the definition of $dx \wedge dy$):
$\omega(\partial_x, \partial_y) = 1$.
$\omega(\partial_y, \partial_x) = - \omega(\partial_x, \partial_y) = -1$.
$\omega(\partial_x, \partial_z) = 0$. Similarly, if we plug in $\partial_z$ into either slot in $\omega$ (no matter what is in the other slot), we get zero, since $dz$ does not appear in $\omega$.
$\omega(\partial_x, \partial_x) = - \omega(\partial_x, \partial_x)$ by the antisymmetry property of forms, so $\omega(\partial_x, \partial_x) = 0$. (By the same reasoning, $\omega(X, X) = 0$ for any $X$ and any two-form $\omega$.)
An exercise in linearity: $\omega(2 \partial_x + \partial_z, 3 \partial_x - 2\partial_y) = -4$ (do you see why?).
Now let's work out $\iota_{\partial_x} \omega$, the contraction of $\omega = dx \wedge dy$ with $\partial_x$. To do so, we need to find what $\iota_{\partial_x} \omega$ is on each of the three basis vectors $\partial_x$, $\partial_y$, and $\partial_z$:
$[\iota_{\partial_x} \omega] (\partial_x) = \omega(\partial_x, \partial_x) = 0$.
$[\iota_{\partial_x} \omega] (\partial_y) = \omega(\partial_x, \partial_y) = 1$.
$[\iota_{\partial_x} \omega] (\partial_z) = \omega(\partial_x, \partial_z) = 0$.
Put succinctly, the above information tells us simply that $$\iota_{\partial_x} \omega = dy.$$
In hindsight, we didn't have to write out all of the above. Instead, we could have seen that $\iota_{\partial_x} \omega = dy$ just by "plucking off" the $dx$ (which "eats" the vector $\partial_x$ we're contracting with) from the front of $dx \wedge dy$ to leave $dy$.
You can apply this somewhat formal and mindless "plucking off rule" to compute other contractions, but you have to be careful with signs. For example, $$\iota_{\partial_y} \omega = \iota_{\partial_y} (dx \wedge dy) = \iota_{\partial_y} (- dy \wedge dx) = -dx.$$ The $\partial_y$ plucks off the $dy$, leaving the $dx$, but we had to move the $dy$ to the first position in order to pluck it off, introducing a minus sign. I would encourage you to check that this is correct using the first (longer) technique above of evaluating $ \iota_{\partial_y} \omega$ on each of $\partial_x$, $\partial_y$, and $\partial_z$.
I hope that if you understand this example, you will be able to do your problem simply by using linearity.