The equality $$T_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$ is false in general, because $T_K(u)$ is closed by definition, whereas $$S_K(u):=K-\mathbb R_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$ can be not closed.

A counterexample is provided by this answer. Indeed, let $$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$ and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.

Also, $w:=(0,1,0)\notin S_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting $$w_t:=v_t-tu$$ for real $t\ge0$ and $v_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v_t\in K$ and hence $w_t\in K-\mathbb R_+ u=S_K(u)$, whereas $w_t\to w$ as $t\to\infty$. $\quad\Box$

The equality $$T_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$ is false in general, because $T_K(u)$ is closed by definition, whereas $$S_K(u):=K-\mathbb R_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$ can be not closed.

A counterexample is provided by this answer. Indeed, let $$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$ and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.

Also, $w:=(0,1,0)\notin S_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting $$w_t:=v_t-tu$$ for real $t>0$ and $v_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v_t\in K$ and hence $w_t\in K-\mathbb R_+ u=S_K(u)$, whereas $w_t\to w$ as $t\to\infty$. $\quad\Box$

The equality $$T_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$ is false in general, because $T_K(u)$ is closed by definition, whereas $$S_K(u):=K-\mathbb R_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$ can be not closed.

A counterexample is provided by this answer. Indeed, let $$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$ and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.

Also, $w:=(0,1,0)\notin S_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting $$w_t:=v_t-tu$$ for real $t\ge0$ and $v_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v_t\in K$ and hence $w_t\in K-\mathbb R_+ u=S_K(u)$, whereas $w_t\to w$ as $t\to\infty$. $\quad\Box$

The equality $$T_K(u)=\{v-tu\colon v\in K,\, t\ge0\}$$ is false in general, because $T_K(u)$ is closed by definition, whereas $$S_K(u):=K-\mathbb R_+ u=\{v-tu\colon v\in K,\, t\ge0\}$$ can be not closed.

A counterexample is provided by this answer. Indeed, let $$K=\big\{(x,y,z)\in\mathbb R^3\colon z\ge\sqrt{x^2+y^2}\big\}$$ and $u=(-1,0,1)$. Then $K$ is a closed convex cone in $\mathbb R^3$ and $u\in K$.

Also, $w:=(0,1,0)\notin S_K(u)$ -- otherwise, we would have $w+tu=(-t,1,t)\in K$ for some real $t$. However, letting $$w_t:=v_t-tu$$ for real $t\ge0$ and $v_t:=(-t,1+1/t,\sqrt{t^2+(1+1/t)^2})$, we have $v_t\in K$ and hence $w_t\in K-\mathbb R_+ u=S_K(u)$, whereas $w_t\to w$ as $t\to\infty$. $\quad\Box$