A key part of the story here is that a $C^*$-algebra whose self-adjoint elements all have finite spectrum must be finite-dimensional. Martin Argerami gave a reference which implies this result by way of several more general results in the context of Banach algebras. However, I decided to write up an account specifically for the $C^*$-algebra case in order to take advantage of the simplifications this context has to offer.
Claim 1: Suppose $A$ is a unital $C^*$-algebra all of whose self-adjoint elements have one point spectrum. Then $A$ is one-dimensional.
Sketch: Use functional calculus to show every self-adjoint element is a multiple of $1_A$, then use the fact that every $C^*$-algebra is the span of its self-adjoint part.
Claim 2: Let $A$ be a $C^*$-algebra in which every self-adjoint element has a finite spectrum and let $p$ be a nonzero projection in $A$ such that the "corner algebra" $pAp$ is not one dimensional. Then, one can write $p=p_1+p_2$ where $p_1$ and $p_2$ are nonzero, orthogonal projections.
Sketch: Working inside $pAp$, whose unit is $p$, the previous claim gives a self-adjoint element whose spectrum is a finite set with at least two elements. Now use functional calculus.
Claim 3: Let $A$ be unital $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then there exist nonzero, pairwise orthogonal projections $p_1,\ldots,p_n$ summing to $1_A$ such that each of the corner algebras $p_iAp_i$ is one-dimensional.
Sketch: If $A$ is not one-dimensional, subdivide $1_A$ into two projections. If either of the new projections does not determine a one-dimensional corner, subdivide again. This process must eventually terminate and give a decomposition of the desired type. Otherwise, we would have an infinite collection of pairwise orthogonal projections in $A$ which could then be used to embed a copy of $c_0(\mathbb{N})$ into $A$ leading to self-adjoint elements of infinite spectrum, contrary to assumption.
Claim 4: Let $p$ and $q$ be nonzero, orthogonal projections in a $C^*$-algebra $A$ satisfying that $pAp$ and $qAq$ are one-dimensional. Then $qAp$ and $pAq$ are either both one-dimensional or both zero.
Proof: Since $(qAp)^* = pAq$, it suffices to look at $qAp$. Fix any nonzero $a \in qAp$. By the $C^*$-identity, $a^*a \in pAp$ and $aa^* \in qAq$ are nonzero. Since $pAp$ is one-dimensional, up to rescaling $a$, we may assume $a^*a=p$. But then $a$ is a partial isometry, so $aa^*$ is also a projection and, being a nonzero projection in $qAq$, is equal to $q$. The identities $a^*a=p$ and $aa^*=q$ show that $x \mapsto ax : pAp \to qAp$ and $x \mapsto a^* x : qAp \to pAp$ are mutually inverse bijections, and the result follows.
Claim 5: Let $A$ be a unital $C^*$-algebra and let $p_1,\ldots,p_n$ be (necessarily orthogonal) projections satisfying $1_A=p_1 + \ldots + p_n$. Then, $A$ is the internal direct sum of the spaces $p_i A p_j$ where $1 \leq i,j \leq n$.
Basically, each element of $A$ can be thought of as a "matrix of operators" with respect to this partition of the identity.
Final Claim: Let $A$ be a $C^*$-algebra all of whose self-adjoint elements have finite spectrum. Then, $A$ is finite-dimensional.
Proof: First unitize $A$ if necessary (which does not affect the spectra of any of its elements). Then combine Claims 3, 4 and 5.
Well, there is one very simple reason: every commutative $C^*$-algebra is isomorphic to $C_0(X)$ for some locally compact Hausdorff space $X$, but not every commutative $C^*$-algebra is isomorphic to $C_b(X)$ for some locally compact Hausdorff space $X$. So the form $C_0(X)$ is not just "prototypical"; it captures every commutative $C^*$-algebra (up to isomorphism).
You can quickly see that not every commutative $C^*$-algebra is isomorphic to an algebra of the form $C_b(X)$ because $C_b(X)$ is always unital. However, this is the only obstruction: every unital commutative $C^*$-algebra is isomorphic to an algebra of the form $C_b(X)$. However, if you restrict to unital algebras, you can do even better: every unital commutative $C^*$-algebra is isomorphic to an algebra of the form $C(X)$ where $X$ is a compact Hausdorff space. Moreover, this $X$ is unique up to homeomorphism and this even extends to a contravariant equivalence of categories between the category of unital commutative $C^*$-algebras and the category of compact Hausdorff spaces. So in the unital case it is better to consider such $C(X)$ as the "prototypical" example.
Best Answer
I don't know how to find the commutant, but I know it is usually strictly bigger than $\ell^\infty(X)$.
For instance let $X=[0,1]$, with the usual topology. Let $E\subset X$ be a non-Lebesgue-measurable set. Let $h=1_E\in\ell^\infty(X)$. If $h\in C_0(X)''$, then by the Double Commutant Theorem there exists a net $\{f_j\}\subset C_0(X)$ such that $f_j\to h$ in the sot. This means that $f_jg\to hg$ in $\ell^2(X)$ for all $g$. That is, $$ \lim_j\sum_{x\in X} |f_j(x)-h(x)|^2\,|g(x)|^2=0,\qquad g\in\ell^2(X). $$ This in particular implies that $f_j\to h$ pointwise. But a pointwise limit of continuous functions is measurable, a contradiction.
So $C_0(X)''\subsetneq\ell^\infty(X)$, which implies $\ell^\infty(X)\subsetneq C_0(X)'$.
As $\ell^\infty(X)$ is maximal abelian, this means that $C_0(X)'$ is non-abelian.