Reviving an old post here.
As I am sure Griffiths and Harris mention in the book, the bundle $T'M$ (or $T^{1,0}M$ in other books) is supposed to be the holomorphic tangent bundle. One way to specify what that is to say that it should act on holomorphic functions, in other words it should consist of derivations mapping holomorphic functions back to holomorphic functions.
Now note that given a holomorphic function $f$ on an open subset of $\mathbb{C}^n$, we have $$ \frac{\partial{f}}{\partial{x_j}}=\frac{\partial{f}}{\partial{z_j}}, \quad \frac{\partial{f}}{\partial{y_j}}=i\frac{\partial{f}}{\partial{z_j}}$$where $\frac{\partial{}}{\partial z_j}$ can either be given by the formula you mention in your post, or simply as the partial differentiation symbol in the $j$-th direction (which makes sence, because the limit in the partial derivative exists by the virtue of $f$ being holomorphic). In this sense it is a pretty natural identification.
Let's just work in $\mathbb{R}^n$ to try to get some geometric intuition. I will directly connect the intuitive definition with the abstract one.
First, let's actually start with $n=3$. In $\mathbb{R}^3$, we can visually a vector at a point $p$ as an arrow starting at $p$, where the direction of the arrow is based on the coordinates of the vector. Visually, we view the tangent plane to $p$ as a plane the touches a surface only at the point $p$. Then, tangent vectors to this surface are all of the "arrows" starting at $p$ that lie in this plane. Note that this requires us viewing our space as being embedded in $\mathbb{R}^3$.
This concept doesn't carry over to manifolds very well. Instead, we will construct an equivalent characterization which does. The tangent space $T_p\mathbb{R}^n$ to $\mathbb{R}^n$ at a point $p\in\mathbb{R}^n$ consists of all arrows starting at $p$. If we view tangent vectors this way then we get an isomorphism between the tangent space and $\mathbb{R}^n$ by sending arrows to column vectors. For now, we'll endow the tangent space with the standard basis $e_1,\cdots, e_n$. Let $p=(p^1,\cdots, p^n)$ be a point and $v=\langle v^1,\cdots, v^n\rangle $ be a vector, all in $\mathbb{R}^n$ (the bracket notation to distinguish a vector from a point). The line through $p$ with direction $v$ is $c(t)=(p^1 +tv^1,\cdots, p^n+tv^n).$ If $f$ is smooth on a neighborhood of $p$, then we can define the directional derivative of $f$ in the direction of $v$ at $p$ to be $$D_v f=\frac{d}{dt}\Big|_{t=0}f(c(t)).$$ Via the chain rule, $$D_v f=v^i\frac{\partial f}{\partial x^i}(p),$$ which is a number, not a function. We can write $$D_v=v^i \frac{\partial }{\partial x^i}\Big|_p,$$ which takes a function to a number. The map $v\mapsto D_v$, which sends a tangent vector to an operator on functions, will give a useful way to describe tangent vectors.
First, we define an equivalence class of smooth functions in a neighborhood of $p$ as follows: consider pairs $(f,U),$ where $U$ is a neighborhood of $p$ and $f\in C^\infty(U).$ We define the relation $\sim$ as $(f,U)\sim (g,V) $ if there exists an open set $W\subset U\cap V$ containing $p$ so that $f=g$ on $W$. An equivalence class of $(f,U)$ is called a germ of $f$ at $p$, and we write $C_p^\infty(\mathbb{R}^n)$ to be the set of all germs of smooth functions on $\mathbb{R}^n$ at $p$, which forms an $\mathbb{R}$-algebra.
Now, given a tangent vector $v$ at $p$, we have $D_v:C_p^\infty\rightarrow\mathbb{R}$. It's not hard to obtain that $D_v$ is linear and satisfies the Leibniz rule: $$D_v(fg)=(D_vf)g(p)+f(p)D_vg.$$ A map $C^\infty_p\rightarrow\mathbb{R}$ that has these two properties is called a point derivation of $C_p^\infty.$ Call this set $\mathcal{D}_p(\mathbb{R}^n)$, which forms a vector space.
Okay, so we've defined a bunch of stuff, but it turns out to be very useful. We have that all directional derivatives at $p$ are derivations at $p$. So, we have a map $\Phi:T_p\mathbb{R^n}\rightarrow \mathcal{D}_p(\mathbb{R}^n)$ given by $v\mapsto D_v,$ as hinted at earlier. By linearity of $D_v$, this is a linear map. It does not take too much work (I can add, if necessary), that this map is actually an isomorphism between $T_p\mathbb{R}^n$ and $\mathcal{D}_p(\mathbb{R}^n)$. Under this isomorphism, we can identify the standard basis $e_1,\cdots, e_n$ for $T_p\mathbb{R}^n$ with the set $\partial/\partial x^1\big|_p,\cdots,\partial/\partial x^n\big|_p.$ That is, if $v$ is a tangent vector, then we can write $v=v^ie_i$ as $$v=v^i\frac{\partial}{\partial x^i}\Big|_p.$$ So, we can define tangent vectors in a geometric way in $\mathbb{R}^n$, and it turns out to be equivalent to the derivation definition, where as you said, tangent vectors are operators (specifically linear functionals on $C_p^\infty$). This definition extends way more naturally to general manifolds than the "arrow" one.
This follows the construction from Tu's "Introduction to Manifolds" fairly closely, which you might like to check out, as a further reference.
Best Answer
To distinguish between points and tangent vectors, let $p=(p_1,...,p_n)\in \mathbb{R}^n$ a point of $\mathbb{R}^n$ and $v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$ a point of the tangent space $\mathbb{R}^n$.
The line through $p=(p_1,...,p_n)\in \mathbb{R}^n$ with direction $v=(v_1,...,v_n)\in T_p(\mathbb{R}^n)$ has parametrization $a(t)=(p_1+tv_1,...,p_n+tv_n)$.
If f is $C^\infty$ in a neighborhood of $p\in \mathbb{R}^n$ and $v$ is a tangent vector at $p$, define the directional derivative of $f$ in the direction of $v$ at $p$ as $$D_vf=\lim\limits_{t \to 0} \frac{f(a(t))-f(p)}{t}.$$
By the multi-variable chain rule, we have $$D_vf=\sum_{i=1}^{n} \frac{da^i}{dt}(0)\frac{\partial f}{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial f}{\partial x^i}(p).$$
Of course in the above notation $D_vf$, the partial derivatives are evaluated at $p$, since v is a vector at $p$. Now, we can define a map $D_v$ (which assigns to every f which is $C^\infty$ the real number $D_v(f)$ ) with the natural way $$D_v=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}(p)=\sum_{i=1}^{n} v_i\frac{\partial }{\partial x^i}\Bigr\rvert_{p}.$$
This map $D_v\in \mathcal{D}_p(\mathbb{R}^n)$ is in fact a derivation at $p$. Finally, you can show that the map \begin{align} \phi :T_p(\mathbb{R}^n) &\to \mathcal{D}_p(\mathbb{R}^n) \\ v &\to D_v \end{align}
is a linear isomorphism of vector spaces (for surjectivity, you can use Taylor's theorem).
So the answers to your questions are:
3) Υes, you can see every tangent vector $v \in T_p(\mathbb{R}^n)$ as a derivation $D_v\in \mathcal{D}_p(\mathbb{R}^n)$ using the isomorphism $\phi$.
2) Since $e_1,...,e_n$ is the canonical basis of $T_p(\mathbb{R}^n)$ and $\phi$ is an isomorphism, then $\phi(e_1),...,\phi(e_n)$ is a basis of $D_v\in \mathcal{D}_p(\mathbb{R}^n)$. But $\phi(e_i)=\frac{\partial }{\partial x^i}\Bigr\rvert_{p}$, hence $\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$ is a basis of the tangent space $\mathcal{D}_p(\mathbb{R}^n)\simeq T_p(\mathbb{R}^n)$.
1) As a result, you can say that the basis of $T_p(\mathbb{R}^n)$ is $\{\frac{\partial }{\partial x^i}\Bigr\rvert_{p}\}_{i=1}^n$. You can say all that because of $\phi$.