Let $\Omega$ be a bounded convex domain. Then the embedding $I_1:C^{k+1}(\overline \Omega) \to C^{k}(\overline \Omega)$ is compact (Adams - Sobolev Spaces, p. 11), where $C^{k}(\overline \Omega)$ represents the space of functions whose partial derivatives up to order $k$ are bounded and uniformly continuous.
Another example of a compact embedding is $I_2:H^1([0,1]) \to L^2([0,1])$.
It is possible to obtain expressions for the spectrum of either of these two embedding operators? I would like to see what eigenfunctions are being lost/diminished (i.e. whose corresponding eigenvalues are close to zero) when we perform an embedding operation?
In order to talk about the spectrum you need an operator in $E$ (for a Banach space $E$), i.e. map from a space into the same space. You may still view the embeddings as unbounded operators in $E=C^{k}$ (resp. $E=L^2$) with domain $D=C^{k+1}$ (resp. $D=H^1$), but then they are not compact anymore (and the spectrum consists of all of $\mathbb{C}$ for trivial reasons).
(1) To avoid confusion we define $T$ to be the unbounded operator define in the Banach space $E=C^{k}$ with domain $D(T)=C^{k+1}$ and $Tx=x$ for all $x\in D(T)$. I claim that $\mathrm{spec}(T)=\mathbb{C}$.
For $\lambda \neq 1$, $(\lambda - T)$ is not surjective, because it's range is contained in $C^{k+1}$ (which is a proper subset of $C^k$), hence $\lambda \in \mathrm{spec}(T)$. And for $\lambda =1$ we have $(\lambda - T) = 0$ which is not injective, hence $1\in \mathrm{spec}(T)$.
(2) Given an arbitrary operator $T$ in $E$ with domain $D(T)$ you can define the graph norm $\vert\cdot \vert_T:=\vert\cdot \vert+\vert T\cdot \vert$ on $D(T)$. Then $(D(T),\vert\cdot \vert_T$) is complete iff $T$ is closed and you can ask whether $(D(T),\vert\cdot \vert_T)\hookrightarrow E$ is compact. Provided $\mathrm{spec}(T)\neq \mathbb{C}$, this implies that $T$ has compact resolvents and if addtionally $T$ is normal you can conclude that its spectrum consists of possibly $0$ and besides that only discrete eigenvalues with finite multiplicity.
But this has nothing to do with the operator $T$ being compact. (With $T$ and $E$ as above, $T$ is not even closed, i.e. $C^{k+1}$ is not complete with respect to the graph norm.)