Integration over the momentum space is implied right? Does it even matter?
From context, yes I would interpret the measure as an integral over momentum space. I am only speculating, but perhaps they did not include the measure to avoid confusion between the $k$ appearing in the path integral measure, and the $k$ that would appear if they wrote the measure in the action explicitly.
And then, my real question is how do I transform from a functional integral over the field to a functional integral over the fourier transform of the field? What is going on in that functional integration measure and how do I go from () to () and vice versa??
Note the integral equation
\begin{equation}
\tilde{\phi}(k) = \int d^4 x e^{i k x} \phi(x)
\end{equation}
has the form of a matrix equation
\begin{equation}
\tilde{\phi}_n = U_{n j} \phi_j
\end{equation}
if we interpret the sum over $j$ as an integral, and where $U_{n j} = e^{i k_n x_j}$ is a unitary matrix.
Therefore, when doing a change of variables in the integration measure from from $\prod_k d \tilde{\phi}(k)$ to $\prod_x d \phi(x)$, we pick up a trivial Jacobian factor $\det U = 1$.
It actually doesn't even really matter that $U$ is unitary; the key point is that $U$ is independent of the field (in other words, we've done a linear change of variables). This is because in the end, we are almost always interested in ratios of partition functions, like $Z[J]/Z[0]$, so an overall field-independent constant will drop out of the final answer.
Putting this together, we can equate the path integrals written in momentum space and real space, since they only differ by a linear change of variables
\begin{equation}
\left(\prod_k \int d \tilde{\phi}_k \right) e^{i \int d^4 k' \mathcal{L}} = \left(\prod_x \int d \phi_x \right) e^{i \int d^4 x' \mathcal{L}}
\end{equation}
The momentum space representation is usually more useful for perturbation theory.
