Obtaining hermitian matrix using Knill and Laflamme condition?

1. I cannot figure out how equations $(1)$ and $(2)$ are equivalent. My humble explanation suggests that for $(1)$, they are equivalent since $E^{dagger}_a E_b$ would cancel out to $I$ and we are left with $langle c_ivert c_j rangle = 0$, since they are orthogonal. For $(2)$ we have $E^{dagger}_a E_b$ would cancel out to $I$ and we are left with $langle c_ivert c_i rangle = langle c_jvert c_j rangle = 1$, since we are projecting a state on itself.

Eq. $1$ and $2$ are not necessarily equivalent, they are just both necessary for a QECC. Eq. $1$ states that orthogonality between codewords is preserved, even if one error $E_{a}$ acts upon one of the codewords, and some other error $E_{b}$ acts upon the other - that way we can always tell two error apart, regardless of what exactly the state encoded in the subspace is. Eq. $2$ takes care of something else: whatever the state encoded in the subspace is, errors occurring on this state must not reveal anything about the state (otherwise we could learn something about the state, thereby destroying quantum information). In other words, the 'symmetric' inner product cannot depend on what exactly the 'current' codeword (or superposition thereof) is!

You can also check out Gottesman's introduction to QECC's (check section $2.2$ on page $5$ and specifically Eq. $(26)$) - he has what I believe to be a clear explanation on why we have these conditions exactly.

2. How can we convert $C_{ab}$ to matrix, what is the matrix dimension and what is the nature of the elements in the cells of this matrix (a binary matrix or elements in $mathbb{C}$ or something else)?

We say that we get a $|E|times |E|$ matrix $C$, where the $(a,b)$-th element is the inner product $langle c_{i}|E_{a}^{dagger}E_{b}|c_{i}rangle$ - Eq. $2$ tells us that it does not matter what codeword $|c_{i}rangle$ we use, as every codeword should give the same result. Generally this matrix is in $C^{|E|times |E|}$, but if ${E_{1}...E_{|E|}}$ is the set of correctable errors, you can view this set as a basis for the space $mathcal{E}$ of correctable errors. As $C$ is Hermitian, there exist a basis of $mathcal{E}$ such that $C$ becomes diagonal with real entries. These entries are not necessarily $1$ (they will be $geq 0$ though, and generally speaking they are $leq 1$). However, these scaling factors are relatively meaningless, and for additive (i.e. stabilizer) codes, in this particular basis the entries are normally $1$.

So, if our set of correctable errors is ${E_{1}...E_{|E|}}$, we get for our matrix $C$: $$ C = begin{bmatrix} langle c_{i*}|E_{1}^{dagger}E_{1} | c_{i*}rangle & langle c_{i*}|E_{1}^{dagger}E_{2} | c_{i*}rangle & cdots & langle c_{i*}|E_{1}^{dagger}E_{|E|} | c_{i*}rangle langle c_{i*}|E_{2}^{dagger}E_{1} | c_{i*}rangle & langle c_{i*}|E_{2}^{dagger}E_{2} | c_{i*}rangle & cdots & langle c_{i*}|E_{2}^{dagger}E_{|E|} | c_{i*}rangle vdots & vdots & ddots & vdots langle c_{i*}|E_{|E|}^{dagger}E_{1} | c_{i*}rangle & langle c_{i*}|E_{|E|}^{dagger}E_{2} | c_{i*}rangle & dots & langle c_{i*}|E_{|E|}^{dagger}E_{|E|} | c_{i*}rangle end{bmatrix} $$ Note that this is slightly different than your comment, since I do not use two separate codewords - if we use different codewords all entries become zero per the first QECC condition. Of course, per the second condition, the codeword $|c_{i*}rangle$ is completely arbitrary.

3. Links to questions two, how would $delta_{ij}$ affect the matrix?

$delta_{ij}$ is there to 'take care' of your equation $1$ - without it, orthogonality between different codewords would not be preserved. All information about what error has happened is encoded into $C$.