The Dual Pairing My understanding from the reading the Wikipedia article on Dual Pairs is that a dual pair is comprised of two vector spaces $X$ and $Y$  over a field $\mathbb{K}$ together with a nondegenerate bilinear mapping $\langle \cdot, \cdot \rangle: X \times Y \rightarrow \mathbb{K}$. 
For $V$ a vector space and it (algebraic) dual $V^{*}$, I have  seen this notation used to effectively represent the evaluation of an element $f \in V^{*}$ at a point $v \in V$ to yield an element of $\mathbb{K}$, i.e., $\langle f, v \rangle = f(v)$ But, from reading the Wikipedia definition, there seems to be no requirement that $X$ and $Y$ are themselves "dual" spaces. So, if we simply supply an inner product $(\cdot | \cdot)$ for $V$ have we effectively induced a "dual paring" between $V$ and itself, i.e., $\langle u, v \rangle = (u | v)$ ?
Finally, I have also seen the Lie derivative of a smooth function $f$ on a manifold $M$ with respect to a smooth vector field on $M$ defined by $L_vf = \langle df, v \rangle$. How should the "dual paring" in this instance be interpreted? Here, $v$ is in the tangent bundle and $df$ is an element of the cotangent bundle so in this case these spaces truly dual in the algebraic sense. Does this notation then just mean to evaluate the differential of $f$ on $v$? A concrete example would be helpful.

The answer to the question in your second paragraph is: yes, an inner product on a real vector space $V$ determines a dual pairing of $V$ with itself. (If the vector space is complex, then no,because the inner product is not linear in one of its variables, only conjugate-linear)
The answer to the question in your third paragraph is: yes, it just means evaluate the differential of $f$ on $v$.
