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• If $w\in \partial \Sigma_x\backslash \partial^1\Sigma_x$, then $D\exp_x\vert_w$ is singular. [Gallot-Lafontaine, Scholium 3.78 or Petersen, Lemma 5.78 ]
• $\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is dense [Sakai, Remark 4.9], see also here.

Proof. On (the open set) $\exp_x(\mathrm{int}\Sigma_x)$ we have $d(x,y) = \vert \exp_x^{-1}(y)\vert^2$ which is clearly a smooth function in $y$. For the converse assume that $d(x,\cdot)^2$ is smooth on some open set $U\subset M$ and note that it suffices to prove $$U \cap \exp_x(\partial \Sigma_x) = \emptyset \tag{1}.$$ Without loss of generality we may assume that $x\notin U$, then also $d(x,\cdot)$ is smooth in $U$ and has a gradient $G\in C^\infty(U;TM)$. Let $\gamma:[0,l]\rightarrow M$ be a length minimising unit-speed geodesic with $\gamma(0)=x$ and $y=\gamma(l)\in U$. Then $d(x,\gamma(t))=t$ and differentiation yields $$\langle G_y, \dot \gamma(l)\rangle = 1. \tag{*} $$ Since $d(x,\cdot)$ is Lipschitz with constant $\le 1$ (triangle inequality) we have $\vert G_y \vert \le 1$. Further $\vert \dot \gamma(l) \vert =1$ and in light of (*) this is only possible if $ G_y = \dot \gamma(l). $ We conclude: $$\text{Length minimising geodesics which start in $x$ don't intersect in $U$.} \tag{2}$$ Now we can prove (1). Assume to the contrary that $U \cap \exp_x(\partial \Sigma_x) \neq \emptyset$. Then by densitity of $\exp_x(\partial^1 \Sigma_x)$ we also have $U \cap \exp_x(\partial^1 \Sigma_x) \neq \emptyset$ and there are $w,w'\in \partial \Sigma_x$ with $w\neq w'$ and $\exp_x(w) = \exp_x(w')\in U$, which is in contradiction to (2).

• If $w\in \partial \Sigma_x\backslash \partial^1\Sigma_x$, then $D\exp_x\vert_w$ is singular. [Gallot-Lafontaine, Scholium 3.78 or Petersen, Lemma 5.78 ]
• $\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is dense [Sakai, Remark 4.9], see also [Klingenberg, Theorem 2.1.12 & 14] as well as here.

Proof. On (the open set) $\exp_x(\mathrm{int}\Sigma_x)$ we have $d(x,y)^2 = \vert \exp_x^{-1}(y)\vert^2$ which is clearly a smooth function in $y$. For the converse assume that $d(x,\cdot)^2$ is smooth on some open set $U\subset M$ and note that it suffices to prove $$U \cap \exp_x(\partial \Sigma_x) = \emptyset \tag{1}.$$ Without loss of generality we may assume that $x\notin U$, then also $d(x,\cdot)$ is smooth in $U$ and has a gradient $G\in C^\infty(U;TM)$. Let $\gamma:[0,l]\rightarrow M$ be a length minimising unit-speed geodesic with $\gamma(0)=x$ and $y=\gamma(l)\in U$. Then $d(x,\gamma(t))=t$ and differentiation yields $$\langle G_y, \dot \gamma(l)\rangle = 1. \tag{*} $$ Since $d(x,\cdot)$ is Lipschitz with constant $\le 1$ (triangle inequality) we have $\vert G_y \vert \le 1$. Further $\vert \dot \gamma(l) \vert =1$ and in light of (*) this is only possible if $ G_y = \dot \gamma(l). $ We conclude: $$\text{Length minimising geodesics which start in $x$ don't intersect in $U$.} \tag{2}$$ Now we can prove (1). Assume to the contrary that $U \cap \exp_x(\partial \Sigma_x) \neq \emptyset$. Then by densitity of $\exp_x(\partial^1 \Sigma_x)$ we also have $U \cap \exp_x(\partial^1 \Sigma_x) \neq \emptyset$ and there are $w,w'\in \partial \Sigma_x$ with $w\neq w'$ and $\exp_x(w) = \exp_x(w')\in U$, which is in contradiction to (2).

• If $w\in \partial \Sigma_x\backslash \partial^1\Sigma_x$, then $D\exp_x\vert_w$ is singular. [Gallot-Lafontaine, Scholium 3.78 or Petersen, Lemma 5.78 ]
• $\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is dense [Sakai, Remark 4.9].

Warning. The denseness of $\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is crucial for the proof of Lemma 1, however Sakai does not prove this result in Remark 4.9, but rather states that It is known that.... To me the statement is not trivial, so I would appreciate if someone could explain this result or show how one can work without the density argument.

• If $w\in \partial \Sigma_x\backslash \partial^1\Sigma_x$, then $D\exp_x\vert_w$ is singular. [Gallot-Lafontaine, Scholium 3.78 or Petersen, Lemma 5.78 ]
• $\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is dense [Sakai, Remark 4.9], see also here.

Proof. On (the open set) $\exp_x(\mathrm{int}\Sigma_x)$ we have $d(x,y) = \vert \exp_x^{-1}(y)\vert^2$ which is clearly a smooth function in $y$. For the converse assume that $d(x,\cdot)^2$ is smooth on some open set $U\subset M$ and note that it suffices to prove $$U \cap \exp_x(\partial \Sigma_x) = \emptyset \tag{1}.$$ Without loss of generality we may assume that $x\notin U$, then also $d(x,\cdot)$ is smooth in $U$ and has a gradient $G\in C^\infty(U;TM)$. Let $\gamma:[0,l]\rightarrow M$ be a length minimising unit-speed geodesic with $\gamma(0)=x$ and $y=\gamma(1)\in U$. Then $d(x,\gamma(t))=t$ and differentiation yields $\langle G_y, \dot \gamma(l)\rangle = 1$. In particular $d(x,\cdot)$ has non-vanising gradient at $\gamma(l)$ and thus it is a submersion in a neighbourhood of $\gamma(l)$. This implies that the orthogonal complement of $\dot\gamma(l)\in T_yM$ is spanned by vectors $\dot c(0)$, where $c:(-\epsilon,\epsilon)\rightarrow M$ are curves with $c(0)=y$ and $d(x,c(s))=\mathrm{const}$. Since $\langle G_y , \dot c(0)\rangle = 0$ for such curves, we have $ G_y = \dot \gamma(l). $ We conclude: $$\text{Length minimising geodesics which start in $x$ don't intersect in $U$.} \tag{2}$$ Now we can prove (1). Assume to the contrary that $U \cap \exp_x(\partial \Sigma_x) \neq \emptyset$. Then by densitity of $\exp_x(\partial^1 \Sigma_x)$ we also have $U \cap \exp_x(\partial^1 \Sigma_x) \neq \emptyset$ and there are $w,w'\in \partial \Sigma_x$ with $w\neq w'$ and $\exp_x(w) = \exp_x(w')\in U$, which is in contradiction to (2).

Proof. On (the open set) $\exp_x(\mathrm{int}\Sigma_x)$ we have $d(x,y) = \vert \exp_x^{-1}(y)\vert^2$ which is clearly a smooth function in $y$. For the converse assume that $d(x,\cdot)^2$ is smooth on some open set $U\subset M$ and note that it suffices to prove $$U \cap \exp_x(\partial \Sigma_x) = \emptyset \tag{1}.$$ Without loss of generality we may assume that $x\notin U$, then also $d(x,\cdot)$ is smooth in $U$ and has a gradient $G\in C^\infty(U;TM)$. Let $\gamma:[0,l]\rightarrow M$ be a length minimising unit-speed geodesic with $\gamma(0)=x$ and $y=\gamma(l)\in U$. Then $d(x,\gamma(t))=t$ and differentiation yields $$\langle G_y, \dot \gamma(l)\rangle = 1. \tag{*} $$ Since $d(x,\cdot)$ is Lipschitz with constant $\le 1$ (triangle inequality) we have $\vert G_y \vert \le 1$. Further $\vert \dot \gamma(l) \vert =1$ and in light of (*) this is only possible if $ G_y = \dot \gamma(l). $ We conclude: $$\text{Length minimising geodesics which start in $x$ don't intersect in $U$.} \tag{2}$$ Now we can prove (1). Assume to the contrary that $U \cap \exp_x(\partial \Sigma_x) \neq \emptyset$. Then by densitity of $\exp_x(\partial^1 \Sigma_x)$ we also have $U \cap \exp_x(\partial^1 \Sigma_x) \neq \emptyset$ and there are $w,w'\in \partial \Sigma_x$ with $w\neq w'$ and $\exp_x(w) = \exp_x(w')\in U$, which is in contradiction to (2).