A rank $(1,0)$ tensor is a map $V^{\ast} \rightarrow \mathbb{K}$, where $V^{\ast}$ is the dual space of a module $V$ defined over $\mathbb{K}$.
How does this relate to the standard "high-school" Euclidean vector? i.e. let $V = \mathbb{R}^2$. A Euclidean vector could then be expressed (by a chart) in terms of a 2-tuple of numbers $(a,b)$. How does this relate back to the abstract definition of a $(1,0)$-tensor?
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A general tensor of type $(p, q)$ is a section of the bundle $TX^{\otimes p} \otimes T^*X^{\otimes q}$ over a (smooth, etc.) manifold $X$. Since Euclidean space is contractible, the bundles $TX$ and $T^*X$ are contractible, and a $(p, q)$-tensor over it just the same as a map to $\mathbb{R}^{p+q}$. In more direct terms, there's a convenient translation map between the fibers of this bundle at any two points, which means we can identify an element of a general fiber with an element of $T_0X^{\otimes p}\otimes T_0^*X^{\otimes q} = \mathbb{R}^{p+q}$. This is not the case on a general manifold; we can try to identify fibers via parallel transport, local triviality, etc., but we can't globally identify fibers because the tangent bundle is not necessarily trivial.
We're identifying $V$ with $V^{\ast\ast}$, which is always possible if $\dim V < +\infty$. We look at $\hat{v} \in V^{\ast \ast}$ acting as $\hat{v}(\phi) \stackrel{\cdot}{=} \phi(v)$, where $\phi \in V^\ast$. Then we say that "$\hat{v} = v$" and there you go: $v$ is acting as a $(1,0)-$tensor.
In smooth manifolds, a tangent vector $v \in T_pM$ can act in functions $f \in C^\infty(M)$ by $v(f) \stackrel{\cdot}{=} {\rm d}f_p(v)$, and this sort of behaves like above. You can take $M = \Bbb R^2$ and $v = (a,b)$ in a chart, for concreteness - it is a directional derivative.