Definition. Let $X\subset \mathbb R^n$ be a locally Euclidean subset. Say it has a tangent space at $p\in X$ if there exists a linear subspace $V\leq \mathbb R^n$ of dimension $\dim_p X$ satisfying the following condition.
There exists a neighborhood $U\subset X$ of $p$ in $X$ for which $$\lim_{\substack{h\to 0\\\text{in }U-p}}\frac{\|\pi_{V^\perp}(h)\|}{\|h\|}=0.$$ Here $U-p$ is the neighborhood of the origin in $X-p$ obtained by translation of $U$ by $(-p)\in X\subset\mathbb R^n$.
Is $V\leq \mathbb R^n$ unique?
The usual argument for the uniqueness of the derivative of a map on an open subset of Euclidean spaces relies on convexity of open balls, so is not applicable here.
No. Let $X\subseteq \mathbb R^3$ be the graph of the function $\sqrt{\|\cdot\|}:\mathbb R^2\to\mathbb R,$ i.e. $\{(x,y,\sqrt[4]{x^2+y^2})\}.$ This is the surface of revolution of a cusp. Since $X$ is a graph, it is $2$-dimensional topological manifold. At $p=(0,0,0),$ any vector space $V$ containing $(0,0,1)$ satisfies your condition; in little-o notation, $x=o(x^{1/4})$ and $y=o(y^{1/4})$ near zero, so $h=(x,y,\sqrt[4]{x^2+y^2})=(0,0,\sqrt[4]{x^2+y^2})+o(\sqrt[4]{x^2+y^2}),$ so the projection onto the orthogonal complement of a space containing $(0,0,1)$ will be negligible compared to $\|h\|.$
You might be interested in the "contingent cone" - I think your $V$ can be taken to be any vector space containing the contingent cone.