It depends on the notation you are using. Let $(e_1,...,e_n)$ be the natural basis for $\mathbb{R}^n$.
One way to understand $\partial/\partial x_i$ is to imagine it as a derivation on the direction $e_i$ calculated at a point $p$. In this case, you can interpret $\partial/\partial x_i$ as the unit vector $e_i$.
To understand it better you should show that any given derivation on $C^\infty(V,\mathbb{R})$ (i.e. an element of $T_p \mathbb{R}^n$) is actually a linear combination of $\partial/\partial x_1, ..., \partial/\partial x_n$.
$\varepsilon$ is a parameter that ensures that $\varphi_j$ is smooth. The smoothness of $\varphi_j$ follows if the factor
$$\prod_{i = 1}^n \rho\biggl(\frac{\psi^i(q) - \psi^i(p)}{\varepsilon}\biggr)\tag{1}$$
is non-zero only on a relatively compact subset of the coordinate patch $U$. By the specification of $\rho$, that factor vanishes outside the subset $V$ of the coordinate patch $U$ corresponding to the cube with sidelength $\frac{3}{2}\varepsilon$ and centre $\psi(p)$, provided that cube is contained in $\psi(U)$. One chooses an arbitrary $\varepsilon > 0$ such that the cube $\prod \bigl[\psi^i(p)-\frac{3}{4}\varepsilon, \psi^i(p) + \frac{3}{4}\varepsilon\bigr]$ is contained in $\psi(U)$, then the function in $(1)$ has compact support in $U$, and $\varphi_j$ is smooth on the whole manifold.
The function $\varphi_j$ is a smooth extension of the coordinate function $\psi^j$ from a small neighbourhood - corresponding to $\prod \bigl( \psi^i(p) - \frac{\varepsilon}{2},\psi^i(p) + \frac{\varepsilon}{2}\bigr)$ - of $p$ to all of $M$. Since apparently the operation of tangent vectors is defined in terms of global smooth functions and not locally, we cannot directly use the coordinate functions $\psi^j$, since those are not necessarily extensible to global smooth functions. Therefore we multiply with a cutoff function to obtain a smooth extension of a restriction of $\psi^j$ to a smaller neighbourhood of $p$.
The relation
$$\biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi} \varphi_j = \delta_{ij}\tag{2}$$
implies linear independence by the linearity of differentiation in the $\frac{\partial}{\partial x^i}$. If we have a linear relation
$$\sum_{i = 1}^n c_i \biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi} = 0,$$
then applying that tangent vector to $\varphi_j$ yields
$$0 = \Biggl(\sum_{i = 1}^n c_i \biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi}\Biggr)\varphi_j = \sum_{i = 1}^n c_i \Biggl(\biggl(\frac{\partial}{\partial x^i}\biggr)_{p,\psi}\varphi_j\Biggl) = \sum_{i = 1}^n c_i \delta_{ij} = c_j.$$
Doing that for all $1\leqslant j\leqslant n$ shows all coefficients $c_i$ are $0$, i.e. the linear independence.
Best Answer
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.