This is a long comment, rather than a complete answer. It shows, for any subset $X$ near $0$ and any dimension of the tangent(s), that if there are two tangent spaces in the sense of the OP then they must have nonzero intersection.
Lemma. Let $\{0\}\neq X\subseteq\mathbb{R}^n$ be any subset that has the origin as an accumulation point. Let $V$ and $V'$ be two nonzero linear subspaces (of arbitrary dimension) of $\mathbb{R}^n$ satisfying the limit conditions
$$\lim_{\stackrel{h\in X}{h\to 0}}\frac{|\pi_{V^\perp}(h)|}{|h|}=0\quad\quad(\star_V)$$
$$\lim_{\stackrel{h\in X}{h\to 0}}\frac{|\pi_{V'^\perp}(h)|}{|h|}=0\quad\quad(\star_{V'}).$$
Then $V\cap V'\neq\{0\}.$
Proof. Assume by contradiction that $V\cap V'=0$. Write $h\in\mathbb{R}^n$ as $h=(x,y)\in V\oplus V^\perp$. Since the $\ell^2$ (usual Euclidean) norm $|h|$ and the norm given by $|(x,y)|_{V,V^\perp}:=\sqrt{|x|^2+|y|^2}$ are equivalent norms on $\mathbb{R}^n$, the condition
$$\frac{|y|}{\sqrt{|x|^2+|y|^2}}\to0\;,$$
for $h=(x,y)\to0$ in a given subset having $0$ as accumulation point, is equivalent to the condition $|y|/|x|\to 0$ for $h=(x,y)\to 0$ in the same subset.
For $\varepsilon>0$ let
$$\mathcal{C}_{V,\varepsilon}:=\{ (x,y)\in\mathbb{R}^n\mid |y|\leq \varepsilon |x|\}.$$
This is a closed cone (invariant under dilations) containing $V$. Likewise, let
$$\mathcal{C}_{V',\varepsilon'}:=\{h\in\mathbb{R}^n\mid |\pi_{V'^{\perp}}(h)|\leq\varepsilon' |\pi_{V'}(h)|\}$$
be the analogous cone containing $V'$.
By the remark on equivalence of norms, condition $(\star_V)$ is easily seen to imply the following: for any given $\varepsilon>0$, there is a $\delta<<1$ such that
$$X\cap B_\delta\subseteq \mathcal{C}_{V,\varepsilon}.$$
Here $B_\delta$ is the ball or radius $\delta$ in $\mathbb{R}^n$ centered at the origin. In the same way, condition $(\star_{V'})$ implies that, for any given $\varepsilon'>0$, there is a $\delta'<<1$ ensuring
$$X\cap B_{\delta'}\subseteq \mathcal{C}_{V',\varepsilon'}.$$
In particular, given two $\varepsilon,\varepsilon'$, for $\delta<<1$ we have $X\cap B_\delta\subseteq \mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}$.
Now, assume by contradiction that neither $V\subseteq V'$ nor $V'\subseteq V$. In what follows we are going to choose $\varepsilon$ and $\varepsilon'$ such that $\mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}=\{0\}$. By the previous remarks, for any $\delta<<1$ we would then have $X\cap B_\delta\subseteq\{0\}$ which is a contradiction because the origin is an accumulation point for $X$.
Let $S_{V'}$ be the (compact) Euclidean unit sphere in $V'$, and let $v_0'$ minimize the quantity $|\pi_{V^\perp}(v')|$ for $v'\in S_{V'}$. Since $V'$ is not contained in $V$, $|\pi_{V^\perp}(v_0')|>0$. If we choose $\varepsilon<|\pi_{V^\perp}(v_0')|/|v_0'|$, we have $v_0'\notin\mathcal{C}_{V,\varepsilon}$ and hence
$$V'\cap\mathcal{C}_{V,\varepsilon}=\{0\}.$$
Let $S_V$ be the Euclidean unit sphere in $V$ centered at the origin, and let $v_0$ minimize the quantity $|\pi_{V'^\perp}(v)|$ on the compact set $S_V\cap\mathcal{C}_{V,\varepsilon}$. Again, $|\pi_{V'^\perp}(v_0)|>0$, and we choose an $\varepsilon'<|\pi_{V'^\perp}(v_0)|/|v_0|$. For this choice of $\varepsilon'$, then, we indeed have
$$\mathcal{C}_{V,\varepsilon}\cap\mathcal{C}_{V',\varepsilon'}=\{0\},$$
and this gives the desired contradiction. $\square$
(Edit: I deleted a corollary because it was not correct)
