First, let me fix some terminology (taken from Lee's book).
Let $M$ be a smooth $n$-dimensional manifold. A coordinate chart on $M$ is a pair $(U, \varphi)$ where $U$ is an open subset of $M$ and $\varphi : U \to \tilde{U}$ is a homeomorphism from $U$ to an open subset $\tilde{U} = \varphi(U) \subset \mathbb{R}^n$. The set $U$ is called a coordinate domain.
Now to the situation at hand. As $M$ is Hausdorff, and $p \neq q$, there are disjoint open sets $U, V \subset M$ with $p \in U$ and $q \in V$. As you note, there is no guarantee that $U$ and $V$ will be coordinate domains. While that is true, we can take them to be coordinate domains without loss of generality. Let $(X, \varphi)$, $(Y, \psi)$ be charts (i.e. $X, Y$ are coordinate domains) with $p \in X$ and $q \in Y$ (note, we do not require $X$ and $Y$ to be disjoint).
Claim: The sets $U' = U\cap X$, $V' = V\cap Y$ are coordinate domains with $p \in U'$ and $q \in V'$.
As $U'$ and $V'$ are open, and the restriction of a homeomorphism to an open subset is a homeomorphism, $(U', \varphi|_{U'})$ and $(V', \psi|_{V'})$ are charts; that is, $U'$ and $V'$ are coordinate domains. As $p \in U$ and $p\in X$, $p \in U'$; likewise, as $q \in V$ and $q \in Y$, $q \in V'$.
So, without loss of generality, we can take $U$ and $V$ to be coordinate domains (if they aren't, pass to subsets $U' \subseteq U$, $V' \subseteq V$ which are, and call these sets $U$ and $V$ respectively).
A derivation at $p\in M$ is in particular a linear map $\partial: C^\infty_p \to \mathbb R$ defined on the set of germs of smooth functions at $p$, and similarly for $q$.
So the sets of derivations at $p$ and $q$ are disjoint simply because they consist of maps with different domains (namely $C^\infty_p$ and $C^\infty_q$). And maps with different domains cannot be equal, as follows from the set-theoretical definition of "map".
Best Answer
We need to study the 2 possibles cases:
Considering $\Pi$, the application, such that:
$\Pi: E \rightarrow M$
If:
Since $M$ is Hausdorff there are $U,V$ disjoint, such that: $ \Pi (x) \in U $, $ \Pi (y) \in V$
Then $x \in \Pi ^{- 1}(U)$ and $y \in \Pi^{- 1}(V)$,
Also, $U,V$ are disjoint
$ U \cap V \neq \emptyset $
$ \Pi ^{- 1} (U) \cap \Pi^{- 1} (V) \neq \emptyset. $
If:
By definition for each point $p$, there exists a $p \in U $ neighbourhood of the point that preserves a fiber differencing:
$ \theta: \Pi^{- 1} (U) \rightarrow U \times R^{ n} $.
Then we have:
$\theta(x) = (p, x^{'})$ $ \theta(y) = (p, y^{'})$
Where $ x^{'}, y^{'} $ are vectors in $ R^{n}$
Since $\theta $ is a diffeormorphism, $ x^{'}\neq y^{'} $.
Also $ R^{n}$ is Hausdorff.
The neighbourhoods $ H, U \in R^{n}$ such that $ x^{'}\in H, y^{'} \in V $
Therefore,
$(p, x^{'}) \in U \times H$
and
$(p, y^{'}) \in U \times V$.
Where $ x \in \theta^{- 1} (U \times H)$ and $ y \in \theta^{- 1} (U \times V) $ are disjoint environments.
We concluded that the tangent bundle $TM$ of a manifold $M $ is Hausdorff.