In general, we can write, using $\mathbf{x} = \pmatrix{x_0 & x_1 & \ldots}$ and $\mathbf{y}=\pmatrix{y_0 & y_1 & \ldots}^\top$
$$\mathbf{x}G\mathbf{y} = \sum_{i=1}^m\sum_{j=1}^n x_iy_jg_{ij}$$
where $G$ is conformable (has the correct dimensions with $m$ rows and $n$ columns.)
Thus for your equation
$$\sum_{k_{1}=0}^{N-L-K}\sum_{k_{2}=0}^{N-L-K}\alpha_{k1}\alpha_{k2}d_{k_{1},k_{2}}(2n)=\delta(n)$$
you have
$$ \mathbf{\alpha_1}D_{2n}\mathbf{\alpha_2} =\delta(n) $$
where
$$\mathbf{\alpha_1} = \pmatrix{\alpha_{k1} & \alpha_{k1} & \ldots}$$
and
$$\mathbf{\alpha_2} = \pmatrix{\alpha_{k1} \\ \alpha_{k1} \\ \vdots}$$
and $D_n = [d_{k_1,k_2}(n)]$ the matrix with elements from $d_{k_1,k_2}$.
In words, the $\alpha_{k1}$ scale the $k_1$ row elements, so it is a left multiply. Similarly, $\alpha_{k2}$ scales the column elements, so it is a right multiply.
This would be what is called a quadratic form if you have $\mathbf{y}=\mathbf{x}^\top$. I have to assume that your problem is not quite the same there as you use separate indices on $\alpha$. But at the same time you use the same $\alpha$ which implies $\alpha_{k1}$ may be the same as $\alpha_{k2}$ but in a different order. This would be true for example if your elements $k_1$ and $k_2$ comprised each element of the same set. If this is the case, with the correct ordering you do have a quadratic form.
Quadratic forms behave nice in one particular manner in that there is implied symmetry. To see this write (using $H=\frac{1}{2}G + \frac{1}{2}G^\top$ and $S=\frac{1}{2}G - \frac{1}{2}G^\top$)
$$G = H + S$$ giving
\begin{align}
\mathbf{x}G\mathbf{y} &=\mathbf{x}\left(H+S\right)\mathbf{y}\\
&= \mathbf{x}H\mathbf{y}+\mathbf{x}S\mathbf{y}\\
& =
\end{align}
and (using $H^\top = H$ and $S^\top = -S$)
\begin{align}
\left(\mathbf{x}G\mathbf{y}\right)^\top &=\mathbf{y}^\top\left(H-S\right)\mathbf{x}^\top\\
&= \mathbf{y}^\top H\mathbf{x}^\top-\mathbf{y}^\top S\mathbf{x}^\top\\
& =
\end{align}
Since $\mathbf{x}G\mathbf{y}$ is a scalar, the two are the same number, and you may take the sum and divide by two. Doing such would cancel the skew symmetric portion $S$ of $G$ again if you have $\mathbf{y}=\mathbf{x}^\top$.
Lets continue from here as if you want to solve $\mathbf{x}G\mathbf{x}^\top$ where $G$ is symmetric. Factor $G=LL^\top$ using the Cholesky factorization.
$$\mathbf{x}LL^\top\mathbf{x}^\top = \mathbf{x}L\left(\mathbf{x}L\right)^\top = \delta(n)$$
I think I will let you take it from here with the hope that your problem is served with the treatment thus given.
