The implication does not work. What it is true is that if a measure admits $k$ Alberti representations, it is absolutely continuous with respect to $\mathcal H^k$. Actually, it is absolutely continuous with respect to the $k$-integral geometric measure. In particular, if the measure is concentrated on a $\mathcal H^k$ $\sigma$-finte set, it is $k$-rectifiable.

A counterexample to your claim is $\mathcal H^1$ restrict to $E=K\times K$ where $K$ is the $1/4$ Cantor set. Indeed $E$ is purely unrectiable.

The implication does not work. What it is true is that if a measure admits $k$ Alberti representations, it is absolutely continuous with respect to $\mathcal H^k$. Actually, it is absolutely continuous with respect to the $k$-integral geometric measure. In particular, if the measure is concentrated on a $\mathcal H^k$ $\sigma$-finte set and admits $k$ Alberti representations, it is $k$-rectifiable.

A counterexample to your claim is $\mathcal H^1$ restricted to $E=K\times K$ where $K$ is the $1/4$ Cantor set. Indeed $E$ is purely unrectiable.