Consider the pairs $(n,k)\in\mathbb{N}^2$ with $0\leq k<2^n$, they form a sequence in lexicographic order. Consider now the functions $f_{n,k}$ which are defined as $0$ in $[0,\frac{k}{2^n}]$, $1$ in $[\frac{k+1}{2^n},1]$, and $2^nx-k$ in $[\frac{k}{2^n},\frac{k+1}{2^n}]$, and suppose there is a set $E\subseteq [0,1]$ and a subsequence of $f_{n,k}$ as in the question.
If we define $A_n=\bigcup\{(\frac{k}{2^n},\frac{k+1}{2^n});(n,k)\text{ is in the subsequence}\}$, then the fact that the subsequence has density $1$ implies that when $n$ tends to $\infty$, the measure of $A_n$ tends to $1$. This implies that there is a point $x\in E$ which is in infinitely many of the $A_n$: for example, take a subsequence $A_{n_i}$ such that $\sum_i(1-m(A_{n_i}))<1$; then $m(\cap_k A_{n_i})>0$), so $E\cap(\cap_k A_{n_i})\neq\emptyset$.
Suppose then that $x\in E\cap(\cap_i A_{n_i})$ and we are given $\varepsilon<\frac{1}{2}$; there is no $\delta$ that satisfies the definition of equicontinuity at $x$: indeed, take $i$ such that $2^{-n_i}<\delta$, then there is some interval $(\frac{k}{2^{n_i}},\frac{k+1}{2^{n_i}})$ contained in $(x-\delta,x+\delta)$ and such that $(n_i,k)$ is in the subsequence. So the image of $(x-\delta,x+\delta)$ under the function $f_{n_i,k}$ is all $[0,1]$, thus it contains values at distance $>\varepsilon$ of $f_{n_i,k}(x)$.
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
Yes. Suppose $f$ is uniformly Lebesgue differentiable (ULD).
Let's first note that $f$ at least has a continuous representative, since the functions $f_r(x) = \frac{1}{2r} \int_{B_r(x)} f(y)\,dy$ are each continuous (by dominated convergence if you like), and by the triangle inequality and the ULD condition, they are uniformly Cauchy and converge a.e. to $f$. Therefore we can replace $f$ by its continuous representative, disregard the null set $E^c$, and assume that the ULD condition holds for all $x$.
Now the idea is that the ULD condition implies that $f(y)$ must be close to $f(x)$ for "most" $y \in B_r(x)$. If $x_1, x_2$ are sufficiently close, then $B_r(x_1), B_r(x_2)$ have a lot of overlap, and so "most" of the $y$ in their intersection have $f(y)$ close to both $f(x_1)$ and $f(x_2)$, so they are close to each other.
To make this precise, let $c$ be a small constant
to be chosen laterless than $\frac{1}{8}$. Now fix $\epsilon > 0$, and fix an $r$ so small that $\frac{1}{2r} \int_{B_r(x)} |f(y)-f(x)| \,dy < c \epsilon$ for all $x$. For arbitrary $x$, let $$ A(x) = \left\{ y \in B_r(x) : |f(y) - f(x)| \ge \frac{\epsilon}{2}\right\} $$ then by Markov's inequality $$ m(A(x)) \le \frac{2}{\epsilon} \int_{B_r(x)} |f(y) - f(x)|\,dy < \frac{2}{\epsilon} \cdot 2r \cdot c \epsilon = 4 c r.$$Now suppose $|x_1 -x_2| < r$. Then we have $m(B_r(x_1) \cap B_r(x_2)) > (2-|x_1-x_2|)r > r$. On the other hand, $m(A(x_1) \cup A(x_2)) \le 8cr$. So
if we choosesince we chose $c < \frac{1}{8}$, then we have $$m\left[(B_r(x_1) \cap B_r(x_2)) \setminus (A(x_1) \cup A(x_2))\right] \ge r - 8cr > 0$$ and hence this set is nonempty, so contains some $y$. We have $y \in B_r(x_1) \setminus A(x_1)$ which means that $|f(y) - f(x_1)| < \frac{\epsilon}{2}$, and similarly $|f(y) - f(x_2)| < \frac{\epsilon}{2}$, so that $|f(x_1) - f(x_2)| < \epsilon$.