How can I compute a presentation of the tangent bundle for a smooth manifold defined by a family of polynomials? Consider a smooth manifold $M$ given by a system of smooth functions
$$
\begin{align*}
f_1 = 0 \\
\cdots \\
f_k = 0
\end{align*}
$$
in $n$ variables. This has the algebraic description as the $\mathbb{R}$-algebra
$$
C^\infty(M) = \frac{C^\infty(\mathbb{R}^n)}{(f_1,\ldots,f_k)}
$$
How can I compute the tangent and cotangent bundle in terms of a similar presentation of algebras?

Here is a running example from the answers:
Consider the circle
$$
C^\infty(S^1) = \frac{C^\infty(\mathbb{R}^2)}{x^2 + y^2 - 1}
$$
Then, the normal vector fields to the manifold should be spanned by the vectors
$$
y\frac{\partial}{\partial x} - x \frac{\partial}{\partial y}
$$
since
$$
\left(y\frac{\partial}{\partial x} - x \frac{\partial}{\partial y}\right)(x^2 + y^2 - 1) = 2xy - 2xy = 0
$$
But, this confuses me since the tangent vector evaluated at (1,0) should be
$$
-\frac{\partial}{\partial y}|_{(1,0)}
$$
which looks like a tangent vector if you draw the picture.

For reference, this can be done by looking at the embedding of $M$ inside $\mathbb{R}^n$, restricting the euclidean tangent bundle to $M$ and cutting out the vector fields which vanish on $M$. This can be explicitly presented as the vanishing of $f_1,\ldots, f_k$ and $\nabla f_1, \ldots, \nabla f_k$ where
$$
\nabla f_i = \frac{\partial f_i}{\partial x_1} \partial_{x_1} + \cdots + \frac{\partial f_i}{\partial x_n} \partial_{x_n}
$$
where the $\partial_{x_j}$'s are the vertical coordinates of $T\mathbb{R}^n \to \mathbb{R}^n$
 A: First of all, in order that  the subset $M\subset \mathbb R^n$ defined by the system $$
\begin{align*}
f_1 = 0 \\
\cdots \\
f_k = 0
\end{align*}
$$be a submanifold you have to assume that the Jacobian matrix $J(f)(x)=(\frac{\partial f_i}{\partial x_j}(x))$ has rank $k$ at every $x\in M$.
If this is the case the tangent space to $M$ at $x$  consists of the $n-k$ dimensional subspace $T_x(M)\subset T_x(\mathbb R^n)= \{x\}\times\mathbb R^n $  defined by: $$T_x(M)=  \{x\}\times \{v\in\mathbb R^n\vert d_x  f_i(v)=0   \quad i=1,\cdots ,k \} $$ where of course  $d_x  f_i(v)=\sum_{j=1}^n \frac{\partial f_i}{\partial x_j}(x)\cdot v_j$
Your example
In your example $n=2, k=1,M=S^1$ and $f_1=f=x^2+y^2-1$ .
Hence for $m=(a,b)$ with $a^2+b^2=1$ we have $d_mf(u,v)=2au+2bv$ and thus $$T_mS^1=\{m\}\times \{(u,v)\in \mathbb R^2\vert au+bv=0\}$$ which can (prudently!) be identified  with the line $au+bv=0$ in $\mathbb R^2$ .
In particular  $T_{(1,0)}S^1=\{(1,0)\}\times \{(u,v)\in \mathbb R^2\vert u=0\}$ which can (prudently!) be identified with $\{0\}\times \mathbb R\subset \mathbb R^2$
A: edit: I write sequence other way around, confused myself and gave wrong answer. Now it should be correct.
Let $M\subseteq \mathbb{R}^n$ be the manifold. We have the following exact sequence of vector bundles:
$$0\rightarrow TM\rightarrow T\mathbb{R}^n_{\mid M}\rightarrow\mathcal{N}_{M/\mathbb{R}^n}  \rightarrow 0$$
and now vector fields on $M$ correspond to restrictions of those vector fields $v$ from $\mathbb{R}^n$ that vanish on $f_1$,...,$f_k$ i.e:
$$v(f_1)=...=v(f_k)=0$$
So that the correct answer is:
$$\langle v_{\mid M} \mid v(f_1)=...=v(f_k)=0\rangle$$
