Edit
I am coming back to this question because although the answer I received cleared the matter a little bit I am still confused by the difference between the 2D and 3D case.
So in 2D case one can define abstractly the normal versor N(s) as the unique vector such that the ordered set $\{T(s), N(s)\}$ forms an orthonormal, positively oriented basis of $\mathbb{R}^2.$
Then one sees that since $\dot{T}(s)=\frac{d}{ds}T(s)$ is orthogonal to $T(s),$ one may take $N(s)$ to be equal to $\frac{\dot{T}(s)}{\|\dot{T}(s)\|}$ for all $s$ such that $\text{det}(T(s),\dot{T}(s))=1$ and equal to $- \frac{\dot{T}(s)}{\|\dot{T}(s)\|}$ for all $s$ such that the determinant is equal to $-1.$ This determines the sign of the curvature at each point $s.$
In 3D case instead, one starts with $T(s)$ and $N(s)=\frac{\dot{T}(s)}{\|\dot{T}(s)\|},$ which amounts to choosing by definition that the curvature must be positive. Then since $B= T \land N$ must be orthogonal to both $T$ and $N,$ those three vectors are linearly independent and thus form an orthonormal basis of $\mathbb{R}^3$.
Question
Is $B$ the unique vector such that $\{ T, N, B \}$ forms a positively oriented basis of $\mathbb{R}^3$?
How can I have an interpretation of the sign of the torsion in a way similar to the one I gave above for curvature?
You haven't been explicit in your definitions. In the plane, you can require that $T,N$ form a right-handed orthonormal basis, and then you find that $dT/ds$ is possibly a negative multiple of $N$ (what is going on here is convexity/concavity). However, in three dimensions, there is no way to define $N$ other than by saying it's the unit vector in the direction of $dT/ds$ (assuming that vector is non-zero, of course). The right-handed orthonormal basis comes in when you add in the binormal vector $B$. It then makes sense to ask if the curve is twisting toward $B$ (positive torsion, by most people's definition) or away from $B$ (negative torsion); here is the sign you were attaching to curvature in the case of a plane curve.