Coordinate basis of tangent space I am reading Carroll, Sean. An Introduction to General Relativity: Spacetime and Geometry. 
In the second chapter, he explains at length the concept of manifolds and tangent vector space at a point on the manifold, with a coordinate chart $x^\mu$. The book explains how the basis of the tangent space is, $$\frac{\partial}{\partial x^\mu} \equiv\partial_\mu$$ This means that any directional derivative along any curve parametrized by $\lambda$ can be written as a linear combination of $\partial_\mu$. So a vector can then be defined as, $$\text{X}=X^\mu \partial_\mu\equiv X^\mu \hat{e}_\mu$$
I get the abstract ideas, somewhat, which are being explained here. But I am having a hard time connecting these ideas to the usual ones taught about vectors in Euclidean geometry. For example, the definition of vectors as, $$\text{X}=X^\mu \partial_\mu$$
sort of looks like an operator, like its 'waiting' for something to act on for it to have a meaningful idea. I am not able to understand the link I am missing to make the connection from these abstract concepts to those in Euclidean geometry where vectors are simply written as $$\vec{V}=V_1 \hat{i}+V_2 \hat{j}+V_3\hat{k}$$.
 A: 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.
