Let $X$ be a normed space over $\mathbb F$, and let $Y$ be a non-trivial subspace. Let $x_0 \in X$ be a non-zero element not in $Y$. Let $Z = \text{span}(x_0, Y)$, and define $f \in Z^\ast$ by the equation $$ f(ax_0) = a \|x_0\|, \ \ f(y) = 0, \ \ \forall a \in \mathbb F, y \in Y. $$ My question is whether this constitutes a valid definition of a linear functional in $Z^\ast$. Is there anything that needs to be checked or further conditions that need to be met for this definition to make sense?
The eventual goal is to extend this map into a linear functional $\bar f \in X^\ast$ which is non-zero on $x_0$ but annihilates $Y$.
My intuition for infinite dimensional spaces is currently lacking, so any advice is appreciated.
I don't think that this is a valid definition for $f$. Every $z \in Z$ has a unique representation $z = ax_0 + y$ with $a \in \mathbb{F}$ and $y \in Y$. In order to define an $f \in Z^\ast$ you have to define what $f(a x_0 + y)$ shall be for all such $z = ax_0 + y$.
If you now define $f(ax_0)$ and $f(y)$, you still do not have defined $f$ for $z$ which contain "parts" of both directions. Take for example $X = \mathbb{R}^2, Y = \operatorname{span} \{ (1, 1) \}$ and $x_0 = (1, 0)$. Then $f(2, 1)$ is not defined, but we have $(2, 1) = (1, 1) + (1, 0) \in Z$.
A suitable definition would for instance be
$$f(ax_0 + y) = a \|x_0\|, \quad a \in \mathbb{F}, \ x_0 \in X, \ y \in Y.$$
Then we have $f \in Z^\ast$.