I am confused about how I can actually go about computing the Symmetric Kronecker Product $ \mathbf{M} \odot \mathbf{N}$. It is defined by the identity:
$$ (\mathbf{M} \odot \mathbf{N}) svec\big(\mathbf{K}) = svec( \frac{1}{2} (\mathbf{N} \mathbf{K} \mathbf{M}^{T} + \mathbf{M} \mathbf{K} \mathbf{N}^{T} )\big)$$
Where $svec(\cdot)$ turns a symmetric matrix $\mathbf{K} = \{k_{ij}\} \in \mathbb{R}^{N}$for instance into a vector of dimension $\frac{N(N+1)}{2}$ via:
$$ svec(\mathbf{K}) = [k_{11}, \sqrt{2}k_{12}, \sqrt{2}k_{13}, \ldots, \sqrt{2}k_{1N}, k_{22}, \sqrt{2} k_{23}, \sqrt{2} k_{24}, \ldots \sqrt{2} k_{2N}, \ldots, k_{NN}]^{T} $$
How can I actually compute $ \mathbf{M} \odot \mathbf{N}$? What is this $\mathbf{K}$ matrix, and where does it come from?
NB: In this answer, I've used "svec" to mean "an enumeration of all the entries of a matrix, reading from right to left, top to bottom." The OP used it to mean "an enumeration of the upper triangle of the matrix", and I missed that point. So as you read this answer, consider all uses of "svec" to be uses of, say, Svec, meaning the all-elements version. (Hat tip to @Tim510 for pointing this out to me!)
You know how we write $$ f(x) = x^2? $$
"What is this $x$, and where does it come from?"
That's the question you're asking.
The formula you've given is saying that there's a matrix (I'll call it $Q$) of size $n^2 \times n^2$, and when you apply it to the length-$n^2$ column vector $v$ that represents a symmetric matrix $K$, the result is another length-$n^2$ column vector, which corresponds to some other matrix, described on the right hand side.
What does the matrix for $Q$ actually look like? Well, to answer that, you need to choose a basis. I'm going to use the basis $$ E_{ij} $$ where $i$ and $j$ run from $1$ to $n$, and $E_{ij}$ has a $1$ in row $i$, column $j$, and zeroes everywhere else. You can see that $svec(K)$ is almost the list of coefficients of $K$ in this basis (but with a few $\sqrt{2}$ factors thrown in). I'm going to be consistent with the $svec$ notation and say that the order for this basis is $$ E_{11}, E_{12}, \ldots, E_{1n}, E_{21}, \ldots, E_{nn} $$
Now this basis is the basis for the set of all $n \times n$ matrices. You'll want to remember that symmetric ones form a subspace of these.
I need one more thing: I'm going to let $m_i$ denote the $i$th row of $M$ (and similarly for $N$); hence $m_i^t$ is also the $i$th column of $M^t$. Clear?
Let's just establish one more thing: what does $E_{ij} M^t$ look like? Well, it selects the $j$th row of $M^t$, and puts in the $i$th row of the result (and the rest of the result is zero). Hold that thought.
OK. Let's see what $$ (\mathbf{M} \odot \mathbf{N}) svec\big(\mathbf{E_{11}}) $$ is. According to the formula, it's \begin{align} (\mathbf{M} \odot \mathbf{N}) svec\big(\mathbf{E_{11}}) & = svec( \frac{1}{2} (\mathbf{N} \mathbf{E_{11}} \mathbf{M}^{T} + \mathbf{M} \mathbf{E_{11}} \mathbf{N}^{T} )\big)\\ \end{align} Now let's ask something a little simpler: what's the $(1, 1)$ entry of the matrix inside the "svec" on the right? Why ask that? Because that'll be the first entry of the column vector produced by $svec$. So we want to know the $(1,1)$ entry of $$ \mathbf{N} \mathbf{E_{11}} \mathbf{M}^{T} + \mathbf{M} \mathbf{E_{11}} \mathbf{N}^{T} $$ The first summand is $N$ times a matrix whose first column is that of $M^t$ , and whose other columns are all zeroes. So the upper left entry is just $n_1$ times $m_1^t$, i.e., it's just $n_1 \cdot m_1$.
The other summand has first entry $m_1 \cdot n_1$. So the upper left entry of the matrix is $$ \frac{1}{2} \bigl( m_1 \cdot n_1 + n_1 \cdot m_1\bigr), $$ and the first entry of the $vvec$ is exactly that number.
OK, let's move on to the $(1, 2)$ entry of \begin{align} \frac{1}{2} (\mathbf{N} \mathbf{E_{11}} \mathbf{M}^{T} + \mathbf{M} \mathbf{E_{11}} \mathbf{N}^{T} ) \end{align} Well, the matrix $$\mathbf{E_{11}} \mathbf{M}^{T}$$ has all zeroes in its second column, so the $(1, 2)$ entry of that piece is $0$; the same argument applies to the right-hand piece. So the $(1,2)$ entry we're looking for is $0$.
The same argument applies all the way up to $(1, n)$.
What about the $(2, 1)$ entry of \begin{align} \frac{1}{2} (\mathbf{N} \mathbf{E_{11}} \mathbf{M}^{T} + \mathbf{M} \mathbf{E_{11}} \mathbf{N}^{T} ) \end{align} The first summand's just the product of the second row of $N$, $n_2$, with the first column of $M^t$, i.e., it's $n_2 \cdot m_1$; the second summand is $m_2 \cdot n_1$. So the $(2, 1)$ entry of our matrix is $$ \frac{1}{2}(m_1 \cdot n_2 + m_2 \cdot n_1 ). $$ In general, we can see that the $(i, j)$ entry of
\begin{align} \frac{1}{2} (\mathbf{N} \mathbf{E_{11}} \mathbf{M}^{T} + \mathbf{M} \mathbf{E_{11}} \mathbf{N}^{T} ) \end{align} is zero unless $j = 1$, and for $j = 1$, the entry is $$ \frac{1}{2}(m_1 \cdot n_i + m_i \cdot n_1 ). $$
And these numbers, all stacked up, form the first column of $$ (\mathbf{M} \odot \mathbf{N}) $$ (almost: there's probably a factor of $\sqrt{2}$ in there somewhere, either once or twice!)
You can now do all the reasoning above, once again, to find the entries of \begin{align} \frac{1}{2} (\mathbf{N} \mathbf{E_{12}} \mathbf{M}^{T} + \mathbf{M} \mathbf{E_{12}} \mathbf{N}^{T} ) \end{align} (note that $E_{11}$ has changed to $E_{12}$ here). And pretty soon, you'll notice a pattern: just as the entries for the $E_{11}$ computation were mostly zeroes, the same holds for the $E_{12}$ computation -- with the zeroes in different places. And the same holds for the $E_{pq}$ computation, for any $p$ and $q$.
Brief pause: I cannot recommend highly enough that you write down a $2 \times 2$ matrix $A$, and a different $2 \times 2$ matrix $B$, and actually do these computations to produce a $4 \times 4$ matrix.
OK. So we've worked out all $n^2$ entries in the first column of the matrix $Q = \mathbf{M} \odot \mathbf{N}$; and (if you've mimicked what I did), for the second column as well, and perhaps even for all the columns.
And the result is something very like this: $Q$ is a block matrix in which the $(i,j)$ $n \times n$ block is something like $a_{ij}B + b_{ij}A$, with some coefficient that may involve some $\sqrt{2}$s.
The exact pattern should be apparent to you if you did the numerical example I suggested above.
TL;DR: actually write out the matrix $Q$ in a standard basis, and a pattern will emerge.