I think here is one example. For each rational in $[0, 1]$ of the form $k/2^{n}$, let $S_{k, n} := \{k/2^n\} \times [-\frac{1}{2^{2n}}, \frac{1}{2^{2n}}]$, and define $S = \cup_{k, n \in \mathbb Z_+} S_{k, n} \cup [0, 1] \times \{0\}$.
Identifying the latter set in the union can be shown that the set of $x \in [0, 1]$ with the liminf in question $0$ for all $v \in \mathbb R^2$ is the set $E$ defined as follows:
Denote by $L_k (x)$ the length of the string of $0$’s or $1$’s beginning at the $k$’th decimal place of the binary expansion of $x$. Set
$$E := \{x \in [0, 1]| \ \limsup_{k \to \infty} L_k (x) - k = +\infty\}.$$
This set $E$ has full measure in $[0, 1]$, whence the set of $x \in S$. with $T_S (x) = \mathbb R^2$ is of $\mathcal H^1$ measure $1$.
Finally we note that $S$ is closed and has $\mathcal H^1$ measure $2$, and so $S$ satisfies the requirements of your problem.