Typically, Lyapunov function assumes $0$ as an equilibrium and require $V(0)=0$. If we wanted to analyze the stability of a nonzero equilibrium point $x_0$, most references asks to do a state transformation $x’=x-x_0$.
Instead, can we / why can’t we require that $V(x_0)=0$ and positive everywhere else?
If it is a nonzero equilibrium point $x_{0}$, you can shift it to a new coordinate. Take this system for example: $$ \dot{x} = - x + x_{0} $$ and introduce the error $$ e = x - x_{0}. $$ Taking the time derivative of $e$ and make the necessary substitutions $$ \begin{aligned} \dot{e} &= \dot{x} - \dot{x}_{0} \\ \dot{e} &= - x + x_{0} - \dot{x}_{0} \\ \dot{e} &= - x + x_{0} \\ \dot{e} &= - \left(x - x_{0}\right) \\ \dot{e} &= - e. \\ \end{aligned} $$ Consider the Lyapunov function candidate, which has the quadratic form such that $V\left(0\right) = 0$ and $V\left(e\right) > 0$ for all $e$ with $e \neq 0$: $$ V\left(e\right) = \frac{1}{2} e^{2}. $$ Now, take the time derivative along the trajectory of the error system $$ \begin{aligned} \dot{V}\left(e\right) &= \frac{dV}{de} \frac{de}{dt} \\ \dot{V}\left(e\right) &= e \dot{e} \\ \dot{V}\left(e\right) &= - e^{2} \\ \end{aligned} $$ The result implies that $\dot{V} < 0$ for all $e$ with $e \neq 0$, and it means $e = 0$ is asymptotically stable. So by logic, when $e \rightarrow 0$, then $x \rightarrow x_{0}$.