What is mixed up is the covariance specification in terms of the ambient space on which the Gaussian process is defined, and the operation that transforms a finite dimensional Gaussian random variable to yield a Wishart distribution.
If $\mathbf{X} \sim \mathcal{N}(0, \Sigma)$ is a $p$-dimensional Gaussian random variable (a column vector)
with mean 0 and covariance matrix $\Sigma$, the distribution of $\mathbf{W} = \mathbf{X} \mathbf{X}^T$ is a Wishart distribution $W_p(\Sigma, 1)$. Note that $\mathbf{W}$ is a $p \times p$ matrix. This is a general result about how the quadratic form
$$\mathbf{x} \mapsto \mathbf{x} \mathbf{x}^T$$
transforms a Gaussian distribution to a Wishart distribution. It holds for any choice of positive definite covariance matrix $\Sigma$. If you have i.i.d. observations $\mathbf{X}_1, \ldots, \mathbf{X}_n$ then with $\mathbf{W}_i = \mathbf{X}_i \mathbf{X}_i^T$ the distribution of
$$\mathbf{W}_{1} + \ldots + \mathbf{W}_n$$
is a Wishart $W_p(\Sigma, n)$-distribution. Dividing by $n$ we get the empirical covariance matrix $-$ an estimate of $\Sigma$.
For Gaussian processes there is an ambient space, lets say for illustration that it is $\mathbb{R}$, such that the random variables considered are indexed by elements in the ambient space. That is, we consider a process $(X(x))_{x \in \mathbb{R}}$. It is Gaussian (and for simplicity, here with mean 0) if its finite dimensional marginal distributions are Gaussian, that is, if
$$\mathbf{X}(x_1, \ldots, x_p) := (X(x_1), \ldots, X(x_p))^T \sim \mathcal{N}(0, \Sigma(x_1, \ldots, x_p))$$
for all $x_1, \ldots, x_p \in \mathbb{R}$. The choice of covariance function, as mentioned by the OP, determines the covariance matrix, that is,
$$\text{cov}(X(x_i), X(x_j)) = \Sigma(x_1, \ldots, x_p)_{i,j} = K(x_i, x_j).$$
Disregarding the choice of $K$ the distribution of
$$\mathbf{X}(x_1, \ldots, x_p) \mathbf{X}(x_1, \ldots, x_p)^T$$
will be a Wishart $W_p(\Sigma(x_1, \ldots, x_p), 1)$-distribution.
The product of the two densities in
$$
p(\boldsymbol{\Lambda_0 | X, \Lambda}, \upsilon, D, \boldsymbol{\Lambda_x}) \propto
\mathcal{W}(\boldsymbol{\Lambda} | \upsilon, \boldsymbol{\Lambda_0})
\mathcal{W}(\boldsymbol{\Lambda_0} |D, \frac{1}{D}\boldsymbol{\Lambda_x}) \\
$$
leads to
\begin{align*}
p(\boldsymbol{\Lambda_0 | X, \Lambda}, \upsilon, D, \boldsymbol{\Lambda_x}) &\propto
|\boldsymbol{\Lambda_0}|^{-\upsilon/2}\,\exp\{-\text{tr}(\boldsymbol{\Lambda_0}^{-1}\boldsymbol{\Lambda})/2\}\\ &\times |\boldsymbol{\Lambda_0}|^{(D-p-1)/2}\,\exp\{-D\,\text{tr}(\boldsymbol{\Lambda_x}^{-1}\boldsymbol{\Lambda_0})/2\}\\
&\propto|\boldsymbol{\Lambda_0}|^{(D-\upsilon-p-1)/2}\,\exp\{-tr(\boldsymbol{\Lambda_0}^{-1}\boldsymbol{\Lambda}+D\,\boldsymbol{\Lambda_x}^{-1}\boldsymbol{\Lambda_0})/2\}\,,
\end{align*}
which does not appear to be a standard density. To keep conjugacy of sorts, the right hierarchical prior on $\boldsymbol{\Lambda_0}$ should be something like
$$
\boldsymbol{\Lambda_0}\sim\mathcal{IW}(\boldsymbol{\Lambda_0} |D, \frac{1}{D}\boldsymbol{\Lambda_x})\,.
$$
Best Answer
As I was getting ready to post this, I was able to answer my own question. In accordance with general StackExchange etiquette I've decided to post it anyways in hopes that someone else who runs into this problem might find this in the future, possibly after running into the same issues with sources that I did. I've decided to answer it immediately so that no one wastes time on it since the solution isn't interesting.
$(\dagger)$ is wrong, because the paper linked to in the discussion was using a different parametrization of the Wishart; this wasn't noticed by the discussants. What we should actually have is $$ \frac{|\Lambda|}{|\Psi|} \sim \chi^2_\nu \chi^2_{\nu - 1} \cdots\chi^2_{\nu - D + 1}. \qquad (\dagger\dagger) $$ After this correction, the two formulas lead to the same answer.
At any rate, I think think $(\dagger\dagger)$ is an interesting relationship.
EDIT:
Following probabilisticlogic's advice we can write $\Lambda \stackrel d = \Psi^{1/2} L L^T \Psi^{1/2}$ where lower triangular $L$ has $N(0, 1)$ elements off the diagonal and $\sqrt{\chi^2_{\nu - i + 1}}, (i = 1, ..., D)$ elements on the diagonal. Taking the determinent of both sides gives $(\dagger\dagger)$ immediately.