I have been trying to prove that, letting $M^2$ be a Riemannian manifold and $X,Y$ two differentiable vector fields on a neighborhood $U$ of $p \in M$, then for any tangent vector $v \in T_p M$ the derivative rule: $v \langle X,Y \rangle = \langle \nabla_v X, Y \rangle + \langle X, \nabla_v Y \rangle$ holds.
But to be honest, I'm having troubles even understanding why $v \langle X,Y \rangle$ is a derivative... and of course, I'm far far away of proving this derivative rule... any help?
That holds if $\nabla$ preserves the metric $g = \langle {}\cdot{},{}\cdot{}\rangle$, i.e. $\nabla g = 0$ (for example, if $\nabla$ is the Levi-Civita connection): in general if $X,Y,Z$ are vector fields on $M$ you have $$\nabla g(X,Y,Z) := (\nabla_X g)(Y,Z)= X(g(Y,Z))-g(\nabla_X Y,Z)-g(Y,\nabla_X Z),$$ so you obtain your formula when $\nabla g = 0$.
If $X$ is a vector field on $M$ and $f$ is a function defined around a point $p$, then $X_p(f)$ is the directional derivative of $f$ at $p$ along the direction given by $X_p$. So, with your notation: $f\colon p \mapsto \langle X,Y \rangle_p$ is a function of $p$. Thus $\nu_p(f)=\nu_p(\langle X,Y \rangle)$ is the directional derivative of $f$ at $p$ along the direction given by $\nu_p$.