It is not sequential. Here is an explicit counterexample.
Notation: $\|\mu\|$ is the total variation norm of $\mu$, and $\bar{B}_r \subset M([0,1])$ is the closed norm ball of radius $r$ centered at $0$. $\delta_x$ is the Dirac measure which places a unit point mass at $x \in [0,1]$.
For each positive integer $n > 0$, let $P_n \subset [0,1]$ be any finite set of size at least $(2n)^{2n}$, and let
$$E_n = \{ n(\delta_{x} - \delta_y) : x,y \in P_n, \, x \ne y\} \subset M([0,1]).$$
Notice that $E_n$ is finite, and that $\|\mu\| = 2n$ for every $\mu \in E_n$.
Set $E = \bigcup_{n=1}^{\infty} E_n$. I claim that 0 is a weak-* limit point of $E$. Let $f_1, \dots, f_m \in C([0,1])$ and let $\epsilon > 0$. I will produce $\mu \in E$ with $\left|\int f_k\,d\mu\right| < \epsilon$ for every $k$. Without loss of generality, assume $\|f_k\|_\infty \le 1$ for all $k$ (otherwise divide $\epsilon$ by $\max_k \|f_k\|_\infty$). Define $F : [0,1] \to [-1,1]^m$ by $F = (f_1, \dots, f_m)$. Choose $n > \max(m, 1/\epsilon)$.
We can cover $[-1,1]^m$ with $(2n^2)^{m}$ cubes of side length $1/n^2$. Since $$|P_n| \ge (2n)^{2n} > (2n)^{2m} = (4n^2)^m > (2n^2)^m$$ by the pigeonhole principle there must exist two distinct $x,y \in P_n$ with $F(x), F(y)$ in the same cube. This means that $|f_k(x) - f_k(y)| \le 1/n^2$ for every $k$. So if we take $\mu = n(\delta_x - \delta_y) \in E_n$, we have $\left|\int f_k\,d\mu\right| \le 1/n < \epsilon$ for every $k$. This proves the claim that 0 is a weak-* limit point of $E$.
Now for any $n$, we know that $\bar{B}_n$ is weak-* closed and disjoint from the finite set $E_n$. The weak-* topology is completely regular, so there exists a weak-* continuous function $G_n : M([0,1]) \to [0,1]$ with $G_n = 0$ on $\bar{B}_n$ and $G_n = 1$ on $E_n$. Set $G = \sum_{n=1}^\infty G_n$. Note that on any ball $\bar{B}_n$, only finitely many terms of the sum are nonzero, so the sum makes sense, and the restriction of $G$ to any ball is weak-* continuous. If $\mu_k$ is a weak-* convergent sequence with limit $\mu$, then by the uniform boundedness principle all the $\mu_k$ and $\mu$ lie in some ball $B$. The restriction of $G$ to $B$ is weak-* continuous, so $G(\mu_k) \to G(\mu)$. Thus we have shown $G$ is weak-* sequentially continuous.
On the other hand, $G$ is not weak-* continuous, since $G(0) = 0$ but $G \ge 1$ on $E$.
(The same uniform boundedness argument shows that $E$ is weak-* sequentially closed, but not weak-* closed since it does not contain 0.)
More generally, we can replace $M([0,1])$ by the dual of any separable Banach space. See the paper:
Humphrey, A. James and Simpson, Stephen G. Separable Banach space theory needs strong set existence axioms. Trans. Amer. Math. Soc. 348 (10), 4231-4255, 1996. Open access full text
Theorem 2.5 of that paper shows that for any infinite-dimensional separable Banach space $X$, there is a countable subset $Z \subset X^*$ which is weak-* sequentially closed but weak-* dense. In particular, $Z$ is not weak-* closed, so the weak-* topology on $X^*$ is not sequential in the usual sense of the word.
To show it is not sequential in your sense: As an intermediate step of their construction, they get a sequence which has $0$ as a weak-* limit point but intersects every ball in only finitely many points; we could proceed as we did above to construct a function $G : X^* \to \mathbb{R}$ which is weak-* sequentially continuous but not weak-* continuous. (Their set $Z$ is formed by translating such a sequence by a countable weak-* dense set.)
This works, because: (1) Your definition of absolute continuity is equivalent to the other standard definition, namely, the condition that $|\mu_g|(A)=0$ implies $|\mu_f|(A)=0$ (your condition implies that $f\in BV$, so $\mu_f$ is well defined).
(2) By a sufficiently general version of the (Lebesgue) differentiation theorem, $\lim_{h\to 0} \mu_h(x,x+h)/|\mu_g|(x,x+h)$ exists for $|\mu_g|$-a.e. $x$ and if $\mu_h\ll |\mu_g|$, then the limit computes the Radon-Nikodym derivative $d\mu_h/d|\mu_g|$. Thus your quotient converges to $(d\mu_f/d|\mu_g| )/(d\mu_g/d|\mu_g|)$; the denominator takes the values $\pm 1$, so the division doesn't make any trouble.
In general, a Radon-Nikodym derivative satisfies $\int f (d\mu/d\nu)\, d\nu = \int f\, d\mu$. This gives the formula from part (2) of your question.
Best Answer
$\newcommand{\B}{\mathcal B}\renewcommand{\S}{\mathcal S}$Note that for any $a$ and $b$ such that $0\le a\le b\le1$ \begin{equation} \mu_h((a,b])=h(b)-h(a)= \left\{ \begin{alignedat}{2} & \mu_f(2(a,b])&&\text{ if }(a,b]\subseteq[0,1/2], \\ & \mu_g(2(a,b]-1)&&\text{ if }(a,b]\subseteq(1/2,1], \end{alignedat} \right. \end{equation} where $2A:=\{2x\colon x\in A\}$ and $2A-1:=\{2x-1\colon x\in A\}$ for any $A\subseteq\mathbb R$. So, for any left-open subinterval $(a,b]$ of the interval $[0,1]$, \begin{equation} \begin{aligned} \mu_h((a,b])&= \mu_h((a,b]\cap[0,1/2])+\mu_h((a,b]\cap(1/2,1]) \\ &=\tilde\mu_h((a,b]) \\ &:=\mu_f(2((a,b]\cap[0,1/2]))+\mu_g(2((a,b]\cap(1/2,1])-1). \end{aligned} \end{equation} The function $\tilde\mu_h$ is a finite signed measure on the semiring (say $\S$) of all left-open subintervals of the interval $[0,1]$, and $\sigma(\S)=\B([0,1])$. So, by the uniqueness of the measure extension (cf. e.g. Proposition 13 of Kisil - Introduction to functional analysis), \begin{equation} \begin{aligned} \mu_h(B)&=\mu_f(2(B\cap[0,1/2]))+\mu_g(2(B\cap(1/2,1])-1) \\ &=\mu_f(2B\cap[0,1])+\mu_g((2B-1)\cap(0,1]) \end{aligned} \end{equation} for all $B\in\B([0,1])$.