$T_P V$ only depends on a neighbourhood of $P\in V$ up to isomorphism I'm having trouble understanding the proof of this statement, this is what my notes say
$T_P V$ only depends on a neighbourhood of $P\in V$ up to isomorphism. More precisely, if $P\in V_0\subset V$ and $Q\in W_0\subset W$ are open subsets of affine varieties, and $\varphi:V_0\to W_0$ an isomorphism taking $P$ into $Q$, there is a natural isomorphism $T_P V_0\to T_Q W_0$. 
Proof. By passing to a smaller neighborhood of $P$ in $V$, we can assume that $V_0$ is isomorphic to an affine variety by $^1$. Then so is $W_0$ and and $\varphi$ induces an isomorphism $k[V_0]\cong k[W_0]$ taking $m_P$ into $m_Q$. 
$^1 V_f=\{P\in V\mid f(P)\neq 0\}$ is isomorphic to an affine variety. 
Specifically 


*

*What smaller neighborhood are we passing to and how does it make $V_0$ isomorphic to an affine variety?

*How do we know the induced isomorphism $k[V_0]\cong k[W_0]$ takes $m_P$ into $m_Q$?


Note: Here $m_P$ is the ideal of $P$ in $k[V_0]$. And the tangent space $T_P V_0=m_P/m_P^2$. Similarly for $Q$, $m_Q$ is the ideal of $Q$ in $k[W_0]$ and $T_Q W_0=m_Q/m_Q^2$. 
 A: *

*Proposition: Let $X$ be an affine variety. Let $p \in X$ and $U \subset X$ be an open set containing $p$. Then there is a neighborhood of $p$ contained in $U$ which is isomorphic to an affine variey. Proof: Suppose that $V(f_1, \ldots, f_n) = U^c$, so $U = \cup D(f_i)$. In particular, there is some $f_i$ so that $f_i(p) \not = 0$ (i.e. $p \in D(f_i)$). $D(f_i)$ is affine (see next paragraph), contains $p$ and is contained in $U$.


But the complement of a hypersurface in affine space is an affine variety, by the following trick: to describe the set $h \not = 0$, add an extra variable $t$, and study the locus $ht = 1$. Then the projection that forgets $t$ induces an isomorphism between $V(h)^c$ and $V(ht) = 1$. (I always think of the example $A^1 \setminus 0$ and $V(xy = 1)$ here.)


*It is because $m_p$ is the ideal of functions vanishing at $p$, so if we map $V_0 \to W_0$, sending $P$ to $Q$, then the functions that pullback to functions vanishing at $P$ were exactly the functions vanishing at $Q$.

