There are quite a few definitions of the tangent space to a manifold. Happily, these are all equivalent in some sense.
One definition that you might like, if you like curves, is the following. Let $M$ be a manifold, and let $P \in M$. Then the tangent space $T_PM$ at $P$ consists of smooth curves $\gamma\colon J \to M$, where $J$ is an open interval in $\mathbf R$ containing $0$, such that $\gamma(0) = P$ under the following equivalence relation: $\gamma_1 \sim \gamma_2$ if there exists a chart $\varphi\colon U \to \mathbf R^n$ around $P$ such that $(\varphi \circ \gamma_1)'(0) = (\varphi \circ \gamma_2)'(0)$. (This will then hold for every chart around $P$, by the chain rule.)
This has the advantage of being nearly obviously the same as your (1) when $M$ is embedded in Euclidean space, yet working in great generality.
Now, how can we connect this to the operators in (2)? Call a linear map $X \colon C^\infty(M) \to \mathbf R$ a derivation at $P \in M$ if it satisfies the Leibniz rule
X(fg) = g(P)Xf + f(P)Xg
for all $f, g$. You can check that in $\mathbf R^n$, for $v \in \mathbf R^n$, the directional derivative gives a derivation at $P$: define $\gamma\colon (-\varepsilon, \varepsilon) \to \mathbf R^n$ by $\gamma(t) = P + tv$. Then
f \mapsto D_v|_P f = (f \circ \gamma)'(0)
is a derivation.
In general, we get a well-defined linear map $T_PM \to D_P$ by sending the class of $\gamma$ to the derivation $f \mapsto (f \circ \gamma)'(0)$. It should be straightforward to show that this association is injective, since we could take $f$ to be a coordinate function. It's less obvious (to me, at least) that it's surjective. But this does follow from the multivariable Taylor's formula. I won't write out a proof here, but you can find one in the case of $M = \mathbf R^n$ (so, proving that every derivation has the form $D_v|_P$) as Proposition 3.2 in Lee's book. From this the case of general $M$ follows.
What I like about the second definition is that it doesn't mention charts or equivalence relations, and it is obviously a vector space. It also generalizes, with only slight modifications, to more algebraic settings.
I apologize for the lack of mental pictures here. To me, the first definition I gave is the most intuitive, and you have this natural (!) map to another space which turns out to be an isomorphism.