In a differential geometry course we were given the following two definitions of a geodesic on a manifold $M$ with $TM$ the tangent bundle on $M$:
Definition 1: Let $V \in \Gamma(\gamma, TM)$ for an unparameterized geodesic $\gamma$. Then $\exists$ $\lambda: M \rightarrow \mathbb{R}$ such that $V^a \nabla_a V^b = \lambda V^b$
Definition 2: Let $V \in \Gamma(\alpha, TM)$ for a parameterized geodesic $\alpha$, then $V^a \nabla_a V^b$ = 0
I expect these are compatible but I am having trouble showing it.
The equation corresponding to an unparametrized geodesic $\gamma=\gamma(t)$ (if you want to think of $\gamma$ as a 1-dimensional manidold, then $t$ is just any local coordinate on $\gamma$) is (your Def.1, $\gamma=(\gamma^1,...,\gamma^d)$ in local coordinates on the $d$-dimensional manifold $M$) $$ \frac{d^2\gamma^i}{dt^2}(t)+\Gamma_{jk}^i(\gamma(t))\frac{d\gamma^j}{dt}(t)\frac{d\gamma^k}{dt}(t)=\lambda(t)\frac{d\gamma^i}{dt}(t),\qquad i=1,...,d. $$ Here and below we use Einstein's convention on contracted indices.
Now let $t=t(s)$ and $\alpha^i(s)=\gamma^i(t(s))$ for an yet unspecified change of parametrization between $t,s$. By the chain rule (denoting for simplicity $\frac{dt}{ds}=\frac{dt}{ds}(s)$ and $\frac{d^2t}{ds^2}=\frac{d^2t}{ds^2}(s)$) $$ \frac{d\alpha^i}{ds}(s)=\frac{dt}{ds}\frac{d\gamma^i}{dt}(t(s)), \qquad \frac{d^2\alpha^i}{ds^2}(s)=\left(\frac{dt}{ds}\right)^2\frac{d^2\gamma^i}{dt^2}(t(s))+\frac{d^2t}{ds^2}\frac{d\gamma^i}{dt}(t(s)). $$ Plugging this into the first equation (multiplied by $(dt/ds)^2$ and evaluated at $t=t(s)$) you get $$ \frac{d^2\alpha^i}{ds^2}(s)+\Gamma_{jk}^i(\alpha(s))\frac{d\alpha^j}{ds}(s)\frac{d\alpha^k}{ds}(s)=\left(\lambda(t(s))\left(\frac{dt}{ds}\right)^2+\frac{d^2t}{ds^2}\right)\frac{d\gamma^i}{dt}(t(s)), $$ and you can now choose appropriately $t=t(s)$ to make the RHS vanish and obtain your Def.2 (of a parametrized geodesic).
Note that parametrized geodesics extremize the action functional, whilst unparametrized geodesics extremize the length functional, and in the second case $\lambda$ is essentially a Lagrange multiplier.