Let $M$ be a $2$-dimensional compact oriented surface in $\mathbb R^3$ with boundary $\partial M$. For any $p\in M \setminus \partial M$ tangent vectors are defined as speed vectors of smooth curves $\Gamma \colon (-1,1) \to M$, $\Gamma(0) = p$ at point $t=0$, i.e. vectors $\dot \Gamma(0)$. They form the linear space $T_p M$ called the tangent space to $M$ at $p$.
Suppose now that $p \in \partial M$. Then if we use the same definition for tangent vectors (i.e. speed vectors at zero of smooth curves $\Gamma \colon (-1,1) \to M$, $\Gamma(0) = p$) we will obtain only vectors, that are tangent to $\partial M$ at point $p$. My question is how to modify the definition of tangent vectors at boundary points to obtain the tangent vectors to $M$ at $p \in \partial M$? Is it possible to define these tangent vectors as speed vectors of curves $\Gamma \colon (-1,1) \to \mathbb R^3$, $\Gamma(-1,0] \subset M$ or $\Gamma[0,1) \subset M$, $\Gamma(0) = p$ at $0$, i.e. vectors $\dot \Gamma(0)$?
I think the simplest way to define tangent vectors is to not use curves at all. Just say that a vector $v$ is tangent to $M$ at point $p$ if $$\operatorname{dist}(p+tv,M)=o(t),\qquad t\to 0^+\tag{1}$$ This agrees with the usual definition at the non-boundary points. At the boundary points the above definition yields halfspace, as in wspin's comment. If you insist on having a linear space, then either
But you can use curves too, by taking one-sided derivative at the point that is mapped to the boundary. Again, the natural way of doing so (curves begin at $p$) leads to half-plane as the tangent plane. Allowing curve that either begin or terminate at $p$ yields the whole plane.
Aside: from the viewpoint of metric geometry, a tangent space is the [pointed] Gromov-Hausdorff limit of rescaled copies of the surface; as such, it is naturally a halfplane at the boundary points. (And quarter-plane at right-angled corners, etc.)