I'm having trouble getting to grips with what a tangent vector really is.
Let $M$ be an $n$-dimensional manifold, and let $\alpha:(-\epsilon, \epsilon) \to M$ be a curve in $M$ through $p$ (i.e. $\alpha(0) = p$). So for $\forall t \in (-\epsilon, \epsilon), \exists \delta \gt 0$ and a chart $(U, \varphi)$ such that $\alpha(t-\delta, t + \delta) \subseteq U$ and $\varphi \circ \alpha$ is a curve in $\varphi(U)$.
If $f$ is smooth function from $U$ to $\mathbb{R}$ then $f \circ \alpha: (-\epsilon, \epsilon) \to \mathbb{R}$. We define the tangent vector to $\alpha$ at $0$ to be $\alpha'(0): f \mapsto (f \circ \alpha)'(0)$. Thus the tangent vector is an operator. I am further aware that tangent vectors can be seen as differential operators evaluated as $0$.
So let $M = \mathbb{R}^3$ and let $\alpha:(-1,1) \to \mathbb{R}^3$ be defined as $\alpha(t) = (\sin (t), \cos (t), t)$. Then $\alpha'(0) = (1,0,1)$. How do we reconcile this with what I intuitvely expect the map to be: $\alpha'(0): f \mapsto \frac{\partial f}{\partial x_1}\Big|_{t=0} + \frac{\partial f}{\partial x_3}\Big|_{t=0}$?
In this case, the correspondence is reasonably clear, but what if we have a general manifold instead?