Understanding of tangent space in $2$ or $3$ dimensions Suppose I am trying to express the tangent vectors in the basis combinations i.e. the partial derivatives w.r.t. coordinates.Let $x^2+y^2-1=0$ i.e. a circle and from my geometrical ideas it has only one tangent vector at any point say $p$ of it.So the tangent space at that point is just a line.But if I think from the partial derivatives $\partial$/$\partial$$x$,$\partial$/$\partial$$y$(that forms the basis of the tangent space) the tangent space at $p$ has dimension two!How does it possible?
Similar problem occurs if I consider the tangent space at any point on the sphere in $3D$,as the tangent space has dimension $2$ (a simple $2D$ plane) but from the partial derivatives it's $3$!There is no doubt that I'm missing some fundamental things.
Please give me some example where the tangent vectors are expressed by the linear combination of the partial derivatives  i.e. basis elements.
 A: The tangent vector to $f(x,y)=0$ or $y=g(x)$ at a point $P(x_0,y_0)$ is not unique indeed they belong to the tangent line at that point which slope is given by the derivative at that point $y'(x_0)$.
In your example:
$$f(x,y)=x^2+y^2-1=0 \implies 2xdx+2ydy=0\implies\frac{dy}{dx}=-\frac{x}{y}$$
thus the/a tangent vector at $(x_0,y_0) \quad y_0\neq0$ is given by
$$\vec v=\left(x_0,-\frac{x_0}{y_0}\right)$$
In the 3D space $z=f(x,y)$the tangent vectors at a point $P(x_0,y_0,z_0)$ belong to the tangent plane at that point:
$$z-z_0=\frac{\partial f}{\partial x}(x_0,y_0)(x-x_0)+\frac{\partial f}{\partial y}(x_0,y_0)(y-y_0)$$
A: Just because you need 2 coordinates to express a subspace doesn't mean that it is 2 dimensional. Let me elaborate.
Here you define a manifold $M^n$ as the level set of some regular value of a function $f$. For example if $f(x,y)=x^2+y^2-1$ then the circle is the level-set of the regular value $f^{-1}(0)$.
Now, the tangent space at a point $p_0=(p_1,p_2,...,p_n)$ is (canonicaly isomorphic) with the kernel of the jacobian of $f$ at that point. Again in the circle the tangent space at $(p_1,p_2)$ is $\ker(2p_1,2p_2)=\{(x,y) \in T_{(p_1,p_2)}\mathbb{R^2}|2xp_1+2yp_2=0\Rightarrow p_2y=-p_1x\}=x\frac{\partial}{\partial x}-\frac{p_1}{p_2}x\frac{\partial}{\partial y}$
$=(x,-\frac{p_1}{p_2}x)$ which of course is a line. The correct equation for the line in $\mathbb{R^2}$ would then be $(p_1,p_2)+\ker(2p_1,2p_2)$.
The problem you encounter is that you see the circle as a submanifold of $\mathbb{R^2}$ (in other words, that it lives in the plane) so each point $(p_1,p_2)$ of the circle already has a 2 dimensional tangent space as a point of $\mathbb{R^2}$. To find the tangent point of the circle you would have to specify the 1-dimensional subspace you want (as we did in the above paragraph).
Final note: This, among, of course ,other reasons, is why we also define manifolds abstractly without assuming an ambient space.
