If we consider the proper time, $\gamma d\tau = dt$, and 4-velocity $v^{\mu}=\tfrac{dx^{\mu}}{d\tau}=(\gamma c,\gamma \vec{v})$ of a particle with velocity $\vec{v}$, then the corresponding space-time interval is being travelled through at constant speed $-c^2$ according to the Minkowski metric $(-,+,+,+)$, i.e. $$ \eta(\mathbf{v},\mathbf{v}) = -\gamma^2c^2 + \gamma^2v^2 = -c^2. $$
But then, in the presence of a gravitational field, such speed is replaced by $g(\mathbf{v},\mathbf{v})$, which is not necessarily equal to $-c^2$ (i.e. not necessarily constant).
So, in the geodesic equation, which has affine parametrization (i.e. constant speed) $$ \dfrac{d^2x^{\mu}}{ds^2} + \Gamma^{\mu}_{\beta\gamma}\dfrac{dx^{\beta}}{ds}\dfrac{dx^{\gamma}}{ds} = 0.\tag{1} $$
is the parameter $s$ not equal to the proper time $\tau$? Is $\tfrac{dx^{\mu}}{ds}$ not equal to the four-velocity?
And what about a free-falling frame of reference? Such a on observer would measure $$ \dfrac{d^2x^{\mu}}{d\tilde{s}^2} = 0. $$
would we write $d\tilde{s}=d\tau$ in such a case?
