The main difference is that in the first case, the functional is defined on the space of continuous functions (on a compact set) whereas in the second case the functional operates on the integrable functions.
For completeness' sake, I add that there also exists a version for continuous functions for which the integration is performed with respect to a measure. This result is much more general than the ones you mentioned, but in your case, the measure theoretic version and the one with a BV-function are essentially the same. The reason for this is that to each signed measure $\mu$ on an interval corresponds a function with bounded variation $g$ (and vice versa) such that the Riemann-Stieltjes-integral of a continuous function $f$ with respect to $g$ coincides with the Lebesgue-integral of $f$ with respect to $\mu$. See this post for some details.
As for the measure-theoretic versions, whether you have to use a real (i.e. signed) measure or a complex measure just depends on whether your functional is real- or complex-valued. This will most often coincide with whether your continuous/integrable functions are real- or complex-valued, since a linear functional is typically defined to be a linear mapping from a vector space into the underlying scalar field, which can only be $\mathbb{R}$ in the case of real-valued functions.
The Riesz theorem for Hilbert spaces is, although named the same, a completely different story. This theorem is about the interplay of continuous functionals and the inner product whereas the other theorems are concerned with the representation of functionals by means of an integral. The deeper similarity of all of them is that they offer characterisations of a dual space.
EDIT: As pointed out by Martin Argerami, the last paragraph is incorrect, I quote:
Any Hilbert space can be represented as $H=L^2(X, \mu)$ for an appropriate measure space $X$, and so the Riesz Representation Theorem says that you can write a bounded functional $\phi$ by$$\phi(f)=\langle f,g\rangle = \int_X f\,\bar g\,d\mu.$$
Therefore, there is a much more immediate similarity between the Riesz theorems.
Given what you're trying to prove, it's probably most convenient to use the version of the Riesz-Markov theorem for positive functionals. This is the content of Theorems 2.14 and 2.17 in Rudin's Real and Complex Analysis.
Below, I have summarised the relevant conclusions from these theorems in the book. (I have simplified the hypotheses somewhat to make the statement easier to digest. Also, the statement in the book is written for $C^{\mathbb C}(X)$ rather than $C^{\mathbb R}(X)$, but the statement does hold for $C^{\mathbb R}(X)$; in fact, the book proves the statement first for $C^{\mathbb R}(X)$, and then deduces the statement for $C^{\mathbb R}(X)$ from this.)
Let $X$ be a compact Hausdorff space, and let $\Psi : C^{\mathbb R}(X) \to \mathbb C$ be a positive linear functional (i.e. $f \geq 0 \implies \Psi (f) \geq 0$). Then there exists a (positive real) regular Borel measure $\mu$ on $X$ such that $$\Psi (f) = \int_X f \ d\mu $$ for all $f \in C^{\mathbb R}(X)$.
Now let's address your question. $X$ is compact Hausdorff, and we have a positive linear functional $\Psi : C_0^{\mathbb R}(X) \to \mathbb R$ such that $\left\| \Psi \right\| = 1$.
First, a quick observation. Since $X$ is compact, $C_0^{\mathbb R}(X)$ is in fact the same thing as $C^{\mathbb R}(X)$.
Now let $\mu$ be the (positive real) regular Borel measure provided by the Riesz-Markov theorem, which satisfies $\Psi(f) = \int_X f \ d\mu$
for all $f \in C^{\mathbb R}(X)$.
If I understood your question correctly, you want to show that this $\mu$ is a probability measure. By construction, $\mu$ is a positive real measure. It remains to check that $\mu(X) = 1$.
To show that $\mu(X) = 1$, observe that the unit function $\mathbf 1_X$ is in $C^{\mathbb R}(X)$, and
$$ \Psi(\mathbf 1_X) = \int_X \mathbf 1_X \ d\mu = \mu(X).$$
Meanwhile,
$$ \left\| \Psi \right\| := \sup_{f \in C^{\mathbb R}(X), \ |f| \leq 1} |\Psi(f) |.$$
But $\Psi$ is a positive linear functional, so $f_1 \geq f_2 \implies \Psi(f_1) \geq \Psi(f_2)$. Thus
$$ \Psi(-\mathbf 1_X) \leq |\Psi(f) | \leq \Psi(\mathbf 1_X) $$
for all $f \in C^{\mathbb R}(X)$ such that $|f| \leq 1$. So
$$ \sup_{f \in C^{\mathbb R}(X), \ |f| \leq 1} |\Psi(f) | = \Psi(\mathbf 1_X).$$
Hence
$$ \mu(X) = \Psi(\mathbf 1_X) = \left\| \Psi \right\| = 1,$$
which shows that $\mu$ is a probability measure.
Best Answer
Let $m$ denote Lebesgue measure. Define a functional $\Phi$ on $C([0,1])$ by $$ \Phi(f)=L\left(x\mapsto \int_0^x fdm\right) $$ This makes sense, because the function $x\mapsto \int_0^x fdm$ is in $C^1$ whenever $f$ is continuous. Note that $\Phi$ is bounded (by $2\|L\|$). By the Riesz Representation Theorem, there is a Radon measure $\mu$ on $[0,1]$ such that $$ \Phi(f)=\int_0^1 f d\mu $$ for all $f\in C([0,1])$. Now, if $f\in C^1$, by the Fundamental Theorem of Calculus, $$ L(f)=L\left(\int_0^x f'dm-f(0)\right)=\Phi(f')-f(0)\Phi(1)=\int_0^1f'd\mu+Kf(0) $$ where $K=-f(1)$ (with $1$ being the constantly $1$ function). I'm performing a slight abuse of notation in writing $\int_0^x f'dm$ for the function $x\mapsto\int_0^x f'dm$.