Concept: "signed" [positive/ negative] curvatures,
Literature overview: I come to see some rules like curvature is positive/ negative based on the turn of the curve (right/ left) (or) based on concavity/ convexity. Can someone help me with the background on how and what determines the sign of a curve?
For example: In the following figure, the magnitude of curvature is the same for both cases. Only the sign varies and how to determine which curvature Fig is positive and which is negative?
For a plane curve the sign of the curvature is tied to (i) the sense of direction of the curve and (ii) the orientation of the plane.
Ad (i): A curve which is lying before us as a static object has no intrinsic sense of direction. This is the case for both curves in your figures. In such a case the curvature is an "unsigned number" and considered $\geq0$. If a curve is presented by a parametric representation $$\gamma:\quad t\mapsto\bigl(x(t),y(t)\bigr)$$ then it obtains a "sense of direction" which corresponds to increasing $t$. At each point $\gamma(t)$ of the curve the tangent vector ${\bf t}(t):=\bigl(x'(t),y'(t)\bigr)$ (assumed $\ne{\bf 0}$) points into the sense of direction of $\gamma$. If $\gamma$ is the graph of a function $y=f(x)$ then the sense of direction of $\gamma$ is the sense corresponding to increasing $x$.
Ad (ii): At each point of $\gamma$ the tangent vector ${\bf t}$ has a polar angle $\theta(t):={\rm arg}\bigl(x'(t),y'(t)\bigr)$. If the tangent vector turns counterclockwise with increasing $t$, i.e., if $\theta'(t)>0$, then the curvature of $\gamma$ is positive at such a point, otherwise it is negative. In fact, the curvature is given by $$\kappa={\theta'(t)\over s'(t)},\qquad s'(t):=\sqrt{x'^2(t)+y'^2(t)}>0\ .$$
