Let us start with some definitions: let $X$ be a compact metric space, abbreviate $C(X) := C(X;\mathbb{R})$, let $L$ be a bounded linear function $L:C(X)\to C(X)$, and equip $C(X)$ with, say, the uniform norm. It follows from Riesz representation theorem that elements of $C(X)^{*}$ can be identified with signed Borel measures, that is given some $A\in C(X)^*$, there exists a unique signed Borel measure $\mu_A$ such that $\forall f \in C(X):A(f) = \int_X fd\mu_A$ (modulo the $\sigma$-algebra details, if my memory serves regarding Rudin's formulation of the theorem).
Given a Borel measure $\mu$ and a Borel measurable function $T:X\to Y$, the push-forward of $\mu$ w.r.t. $T$ is usually given as $T_{*}\mu(B) := \mu(T^{-1}(B)), B\subset Y$.
Lastly, let us remember that if $X, Y$ are Banach spaces (I can't recall whether this could be loosened to the case that $X$ or $Y$ is just a normed space), the adjoint of $A\in \mathcal{B}(X, Y)$ is the unique operator $A^{*}\in \mathcal{B}(Y^{*}, X^{*})$ with $\mathcal{B}(X, Y)$ the set of bounded linear operators from $X$ to $Y$.
Edit: It seems that a push-forward equality is not really possible, at least as per the given definition of a push-forward, since $(f\circ L)(x)$ is not a priori defined, if $L:C(X)\to C(X), f\in C(X)$. However, let us proceed with a equality of interest by defining the pull-back of $\mu$ w.r.t. $T$ as $T^*\mu(B) = \mu(T(B))$.
Now, given the stated constructs, does it then follow that the adjoint operator $L^{*}$ of an $L\in\mathcal{B}(C(X), C(X))$ is just the pull-back (???) of $L$ on the elements of $C(X)^*$ in the sense that
$$\forall \mu\in C(X)^{*}, f \in C(X): L^{*}(\mu)(f) = \int_X f(x)d(L^{*}(\mu))(x) = \int_X (L\circ f)(x)d\mu(x)$$
If this is true, how would you go about and prove it? As of writing I do not clearly see what the overall strategy would be, other than approximating indicator functions of Borel sets with some sequences of continuous functions $(f_n)_{n}$.