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Quantum operations can be represented in an elegant form known as the operator-sum representationn, namely
$\mathcal{E}(\rho)=\sum_kE_k\rho E_k^\dagger$, where $E_k=\langle e_k|U|e_0\rangle$, the operators $\{E_k\}$ are known as operation elements for the quantum operation $\mathcal{E}$, also called the Kraus operator, but why this form $E_k=\langle e_k|U|e_0\rangle$ not a number as a matrix element?
Where did you get the expression $E_{k} = \langle e_{k} | U | e_{0} \rangle$ from? It seems to me that this comes from a Stinespring dilation, and in that case you need to be careful with the spaces all of these operators are acting upon.
$$ E_{k} = \langle e_{k} |U|e_{0}\rangle $$ is shorthand notation for $$ E_{k} = \big(I_{\mathcal{H}}\otimes \langle e_{k} |_{\mathrm{env}}\big)U\big(I_{\mathcal{H}}\otimes |e_{0}\rangle_{\mathrm{env}}\big), $$ where $\mathcal{H}$ is the original input Hilbert space and $\mathrm{env}$ indicates some environment Hilbert space. $U$ acts on the composition of both these spaces. Therefore, the dimensions only line up if you associate the identities on $\mathcal{H}$ with the basis elements $|e_{k}\rangle$ of the environment in the expression for the Kraus operators.
For the sake of readability, suppose that the operation $\mathcal{E}$ maps from a Hilbert space to itself: $\mathcal{E} : \mathcal{H} \rightarrow \mathcal{H}$. That means that we expect $E_{k}: \mathcal{H} \rightarrow \mathcal{H}$, too, which indeed can be represented by matrices, and not scalars.
You have written down a Stinespring dilation of the channel $\mathcal{E}$. Basically, the idea is that any channel (i.e. CPTP map) acting on a Hilbert space $\mathcal{H}$, is equivalent to the following:
The equivalence is then manifest if we write this out explicitly for a particular input state $\rho_{\mathrm{in}} \in Ops(\mathcal{H})$, a state in the space of operators on $\mathcal{H}$: $$ \begin{split} \mathcal{E}(\rho_{\mathrm{in}}) &= \mathrm{tr}_{\mathrm{env}} \big[U(\rho_{in} \otimes |0\rangle_{\mathrm{env}}\langle0|_{\mathrm{env}})U^{\dagger}\big] \\ &= \sum_{k} \big(I_{\mathcal{H}}\otimes \langle e_{k} |_{\mathrm{env}}\big)\big[U(\rho_{in} \otimes |0\rangle_{\mathrm{env}}\langle0|_{\mathrm{env}})U^{\dagger}\big]\big(I_{\mathcal{H}}\otimes |e_{k} \rangle_{\mathrm{env}}\big) \\ &= \sum_{k} E_{k} \rho_{\mathrm{in}} E_{k}^{\dagger}, \end{split} $$ if we define $$ E_{k} = \big(I_{\mathcal{H}}\otimes \langle e_{k} |_{\mathrm{env}}\big)U\big(I_{\mathcal{H}}\otimes |e_{0}\rangle_{\mathrm{env}}\big). \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) $$ Above, I have used the fact that tracing over a subsystem is the same as summing over (the sandwich product of) a basis for that space, while leaving the rest unaffected.
Taking a closer look at $E_{k}$, we see that it maps elements from $\mathcal{H}$ to $\mathcal{H}$, as we expect it to.
The expression you wrote down is not necessarily incorrect, but it can be ambiguous: it's shorthand notation for the expression $(1)$, if you realize that the $|e_{0}\rangle$ and $\langle e_{k}|$ are elements of $\mathcal{H}$ and its dual, whereas $U$ is an element of $Ops(\mathcal{H} \otimes \mathcal{H}_{\mathrm{env}})$. I have just been more explicit, by writing out every $I_\mathcal{H}$ that pops up.
