We know that the tangent plane $T_pS$ of $S$ at $p$ is a vector space of dimension $2$. We ask here what is the vector structure of $T_pS$. Indeed, let $v,w\in T_pS$ and let $\alpha, \beta: I\rightarrow S$ be two curves such that $\alpha(0)=\beta(0)=p$ and $\alpha'(0)=v$ and $\beta'(0)=w$. Since $T_pS$ is a vector space, then $v+w\in T_pS$. We ask:
Question. What is the curve on $S$ whose tangent vector is $v+w$?
A first attempt would be $\alpha+\beta$, because $(\alpha(t)+\beta(t)'(0)=v+w$. Of course, this is completely wrong because $\alpha(t)+\beta(t)$ is a curve in ${\mathbb R}^3$ which it is not included at $S$! To find the right curve $\gamma:I\rightarrow S$ such that $\gamma(0)=p$ and $\gamma'(0)=v+w$ we have to come back to the moment where we proved that $T_pS$ is a vector space. Recall that $$T_pS=(dX)_q({\mathbb R}^2),\ q=X^{-1}(p),$$ where $X$ is a parametrization around $p$, and $(dX)_q:{\mathbb R}^2\rightarrow {\mathbb R}^3$ is the derivative of the map $X:U\subset {\mathbb R}^2\rightarrow{\mathbb R}^3$. The above identity says us that $v+w$ is the sum of two vectors in ${\mathbb R}^2$. Thus the idea to find $\gamma$ is: first, compute the preimages of $v$ and $w$, namely, $\bar{v}$, $\bar{w}$, compute $\bar{v}+\bar{w}$, then take a curve passing $q$ with tangent vector $\bar{v}+\bar{w}$, and finally, consider the image of this curve by the parametrization $X$.
Let $\bar{\alpha}(t)=X^{-1}(\alpha(t))$, $\bar{\beta}(t)=X^{-1}(\beta(t))$. Then $\alpha(t) = X(\bar{\alpha}(t))$, $\beta(t)=X(\bar{\beta}(t))$. Thus $$v=(dX)_q(\bar{\alpha}'(0)),\ w=(dX)_q(\bar{\beta}'(0)).$$ Here we obtain $\bar{v}= \bar{\alpha}'(0)$ and $\bar{w}= \bar{\beta}'(0)$. Consider $\bar{v}+\bar{w}$ placed at $q$ and consider a curve passing $q$ with this tangent vector: if suffices $$\sigma:I\rightarrow U,\ \sigma(t)=q+t(\bar{v}+\bar{w}),$$ where it is immediate $\sigma'(0)=\bar{v}+\bar{w}$.
Here we find the key of the vector structure of the tangent plane $T_pS$: of course, we can not sum the curves $\alpha$ and $\beta$, neither the curves $\bar{\alpha}$ and $\bar{\beta}$ (in this case, outside of $U$). We take another curve that represents the (tangent) vector $\bar{v}+\bar{w}$. In this case, it is enough a straight-line and here we use strongly that $U$ is an open set of ${\mathbb R}^2$, because at least around $t=0$, the straight-line $\sigma$ is included in $U$.
The last steps are now easy. Consider $\gamma(t)=X(\sigma(t))$, which is a curve on $S$ with $\gamma(0)=X(\sigma(0))=X(q)=p$. Then using the chain rule, we have $$\gamma'(0)=(dX)_{\sigma(0)}(\sigma'(0))= (dX)_q(\bar{v}+\bar{w})=(dX)_q(\bar{v})+(dX)_q(\bar{w})=v+w.$$
Similarly, we can find the curve that represents the tangent vector $\lambda v$, where $\lambda\in{\mathbb R}$.
