Realizing any vector in tangent space as tangent vector Let $S$ be a surface in $\mathbb{R}^3$.
Let $U\subseteq \mathbb{R}^2$ be open (connected) and $\sigma:U\rightarrow S$ be a homeomorphism into $\sigma(U)\subset S$, with $\sigma(U)$ open in $S$, and $\sigma$ is smooth map. 
Assume that $\sigma_x,\sigma_y$ are independent for every point in $U$ (i.e. for every $p=(a,b)\in U$, the partial derivatives of $\sigma$ at $p$ are independent vectors in $\mathbb{R}^3$. [This means $\sigma$ is regular].
Now fix $(x_0,y_0)\in U$ and $v_1=\sigma_x(x_0,y_0)$, $v_2=\sigma_y(x_0,y_0)$. 

It can be shown algebraically that every vector $\alpha v_1 + \beta v_2$ is tangent to surface $S$ at point $\sigma(x_0,y_0)$. Define for small $t$,
  $$\gamma(t)=\sigma(x_0+\alpha t, y_0+\beta t).$$
  This is a curve in $S$ and $\gamma(0)=\sigma(x_0,y_0)$; also at $t=0$ we can see that the tangent to curve (hence surface) $\gamma$ is $\alpha \sigma_x(x_0,y_0)+\beta\sigma_y(x_0,y_0)$. 

I did not see this description pictorially what $\gamma(t)$ expresses? For simple example of sphere, what is this $\gamma$? Can one give some pictorial explanation of above paragraph, with simple example?

Ref. Elementary Differential Geometry - Pressley [New edition], Prop. 4.2
 A: Every vector in the tangent space of a surface at a point is the velocity vector of a curve on the surface that passes through that point. The curve $\gamma$ here is chosen in order to show that every linear combination $\alpha v_1 + \beta v_2$ of the two vectors $v_1$ and $v_2$ also lies in the tangent space, since it is the velocity vector of the curve $\gamma(t)=\sigma(x_0+\alpha t, y_0 + \beta t)$. The vectors $v_1$ and $v_2$ are the vectors tangent to the coordinate lines of the parametrisation.
Here is a picture of a sphere where two coordinate lines at a point are drawn in blue. The red curve is a example of a curve $\gamma$. 

A: The curve $\gamma(t) = \sigma(x_0 + \alpha t, y_0 + \beta t)$ is a curve that lies on the surface $S$, and passes through $\sigma(x_0,y_0) \in S,$ since
$\gamma(0) = \sigma(x_0,y_0),$ and it has $\alpha v_1 + \beta v_2$ as its tangent vector at $0$, since by the chain rule,
$$ \dot\gamma(0) = (\sigma_x(x_0+\alpha t,y_0+\beta t) \alpha + \sigma_y(x_0+\alpha t, y_0+\beta t)\beta)\mid_{t=0} = \alpha\sigma_x(x_0,y_0) + \beta\sigma_y(x_0,y_0) = \alpha v_1+\beta v_2. $$
More specifically, you can take the unit sphere $S^2$ parametrized by
$$ \sigma(u,v) = (\cos u \cos v, \cos u \sin v, \sin u). $$
The partial derivatives are
\begin{align}
\sigma_u &= (-\sin u \cos v, -\sin u \sin v, \cos u)\\
\sigma_v &= (-\cos u \sin v, \cos u \cos v, 0).
\end{align}
Consider the point $p=(u_0,v_0) =(0,0)$ in this parametrization, which corresponds to $(1,0,0)$ in Cartesian coordinates. The partial derivatives at $p$ simplify to
\begin{align}
\sigma_u &= (0, 0, 1), \quad \sigma_v = (0, 1 , 0).
\end{align}
Hence $T_pS$ is just the $yz$-plane (as you would expect). As you mentioned yourself, every vector in $T_pS$ can be written as $\alpha \sigma_u + \beta\sigma_v$. For the sake of simplicity, let $\alpha = 0, \beta  = 1$. Then your curve $\gamma$ becomes
$$ \gamma(t) = \sigma(u_0 + \alpha t, v_0 + \beta t) = \sigma(0,t) = (\cos t, \sin t, 0). $$
This curve is just a circle along the equator passing through $(0,0,1)$ and having $(0,1,0) = \sigma_v = \alpha\sigma_u + \beta\sigma_v$ as its tangent vector (i.e. it is moving east).
If we instead put $\alpha = 1, \beta = 0$, we get
$$ \gamma(t) = \sigma(u_0 + \alpha t, v_0 + \beta t) = \sigma(t,0) = (\cos t, 0, \sin t). $$
This curve is a circle along the meridian passing through $(0,0,1)$ and having $(0,0,1) = \sigma_u = \alpha\sigma_u + \beta\sigma_v$ as its tangent vector (i.e. it is moving south).
