Solve the following optimization problem graphically, draw the tangent cone and determine the linearized feasible directions. $$\min_{x\in\mathbb{R}^2}-x_1$$ $$s.t. x_2-(1-x_1)^3\leq 0, -x_1\leq0, -x_2\leq0 $$
So the minimum is $\bar x =(1,0)$ and the green line is the tangent cone. I determined the linearized feasible directions (red) by taking $g_1(x):=-x_1$ (inactive constraint), $g_2(x):=-x_2$ and $ g_3(x):=x_2-(1-x_1)^3$: $$\nabla g_2(\bar x)=(0,-1)^T, \nabla g_3(\bar x)=(0,1)^T,$$ so $$L(F,\bar x)=\{s\in\mathbb{R}^2: \nabla g(x)_i^Ts\leq 0 \forall i\in\{2,3\}\}=\{s\in\mathbb{R}^2:-s_2\leq0, s_2\leq 0\}=\{s\in\mathbb{R}^2: s_2=0\},$$ the $x_1$ axis.
The tangent cone is not equal to linearized feasible directions. Is this correct?
The tangent cone may be different from the set of linearized feasible directions, but it is always included in it.
However, if the set of active constraint gradients is linearly independent, the equality holds.
Here, the set of active constraint gradients is $\{(0, -1)^T, (0, 1)^T\}$ which is linearly dependent, so not to have equality is not a problem. Nevertheless, the tangent cone should be included in $L(F, \bar{x})$, which is not the case in your graphic.
Your example is a special case where, even though the set of active constraint gradients is linearly dependent, the tangent cone equals the set of linearized feasible directions: $T(\bar{x})=\{ s \in \mathbb{R}^2 \colon s_2=0\} = L(F, \bar{x}).$