Yes, I think you translated the condition correctly. The easiest way for me is to remember this ring is anti-isomorphic (via transposition) to the one in the link you gave.
If $I_1=\{0\}$, then $I_2$ can be any submodule of $K\oplus K$.
If $I_1=K$, then $I_2$ has to contain $\{0\}\oplus K$, so it has to be of the form $K\oplus K$ or $\{0\}\oplus K$.
These last two correspond to
$$\begin{pmatrix} K & 0\\ K & 0 \end{pmatrix}$$
and
$$\begin{pmatrix} K & 0\\ K & K \end{pmatrix}$$
and the ones of the first form look like
$$L_W=\left\{\begin{pmatrix} 0 & 0\\ b & c \end{pmatrix}\middle|\, (b,c)\in W\right\}$$ for a given subspace $W< V$.
Note that if you pick any $1$-dimensional subspace $W$, that is automatically going to make a minimal left ideal. The first two in the list are obviously not minimal because they contain the left ideal of strictly lower triangular matrices.
We can show all minimal left ideals are isomorphic to $L_{K\oplus 0}$. Suppose $(a,b)$ is a nonzero element of $L_W$ where $W$ is one dimensional. If $b=0$ then obviously there is nothing to do. If $b\neq 0$, then
$\begin{pmatrix}0&0 \\ b^{-1}&0\end{pmatrix}$ defines a left $R$ module transformation from $L_W\to L_{K\oplus 0}$ by right multiplication, which is an isomorphism because they're both one dimensional.
So look at what you have:
$$\begin{pmatrix} K & 0\\ K & K \end{pmatrix}\cong \begin{pmatrix} K & 0\\ K & 0 \end{pmatrix}\oplus \begin{pmatrix} 0 & 0\\ 0 & K \end{pmatrix}$$, so both of these pieces are projective, being summands of a free left module.
Then, all of the $L_W$ with one-dimensional $W$ are isomorphic to the second factor, so they are projective as well. If $W$ is two-dimensional, then $L_W=\begin{pmatrix} 0 & 0\\ K & K \end{pmatrix}$ which is obviously a direct sum of two minimal left ideals, and thus projective (since they are.)
So, all left ideals are projective.
Best Answer
For any right ideal $I\subseteq R$, let $I_A\subseteq A$ denote the right ideal formed from the top left entries of elements of $I$. Similarly let $I_B\subseteq B$ denote the right ideal formed from the bottom right entries of elements of $I$.
Then a right ideal $I\subseteq R$ is superfluous if and only if the ideals $I_A\subseteq A, I_B\subseteq B$ are both superfluous.
Proof:
If $I\subseteq R$ is a superfluous ideal, then we know that $I_A$ is superfluous, as otherwise we would have $J_A+I_A=A$ for some proper ideal $J_A\subset A$. Then the ideal:
$$J=\left\{\left(\begin{array}{cc} a&m\\0&b \end{array}\right) |a\in J_A, b\in B, m\in M \in M \right\}, $$ is a proper ideal of $R$, but $I+J=R$, contradicting that $I$ was superfluous.
Thus we know that $I_A$ is superfluous, and the same argument gives that $I_B$ is superfluous.
Conversely suppose that $I_A,I_B$ are superfluous. We will show that $I$ is superfluous.
If $I+J=R$ then the top left entries of elements of $J$ form an ideal $J_A\subseteq A$, and $I+J_A=A$ so $J'=A$. Similarly the bottom right entries of elements of $J$ are all of $B$. We conclude that $J=R$.
It may be helpful to note that in the above proof we use that if an ideal $K\subseteq R$ contains elements $$\left(\begin{array}{cc} 1&m\\0&b \end{array}\right),\qquad\left(\begin{array}{cc} a&m'\\0&1 \end{array}\right) $$ then $K=R$. This follows from the fact that right multiplication by an upper triangular matrix can zero out either column, leaving the other column the same.