The definition of geodesics is not exactly what you stated. A curve is a geodesic if it minimizes locally the distance between points, not globally. But this is not so important.
In the case where $M$ is a submanifold of $\mathbb{R}^n$ with the induced metric, that is, $M \subset \mathbb{R}^n$ and if $u,v \in T_pM$, that is $g_p(u,v) = \langle u,v\rangle$ the usual scalar product, then a smooth curve $\gamma : I \mapsto M$ is a geodesic if and only if its acceleration is orthogonal to $M$, that is, for all $t\in I$, $\gamma''(t) \perp T_{\gamma(t)}M$.
If $(M,g)$ is an abstract riemannian manifold, there is no ambiant space to state that the acceleration of a curve is orthogonal to $M$. In fact, there is nothing that can define the acceleration on an abstract manifold! Even if smoothness is well-defined, second derivative of a function is not well-defined as it is closely coordinate-dependant. So there is no possibility to talk about high order differentiation.
The key is that if $M$ is a riemannian manifold, there exists an intrinsic object call the Levi-Civita connection or covariant derivative, denoted by $\nabla$ or $D$ usually, that allows you to talk about acceleration. It is an algebraic objet that allows you to derive vector fields in the direction of vectors fields. In case $M\subset \mathbb{R}^n$ with the induced metric, the Levi-Civita connection is the orthogonal projection onto the tangent space of the derivative: if $v$ is a vector field and $\gamma(t)$ is a curve, the covariant derivative of $v$ along $\gamma$ is
$$ \nabla_{\gamma'(t)}v(t) = {v(\gamma(t))'}^{\perp} $$
In the general setting, one define the acceleration of a curve $\gamma$ to be the vector field along $\gamma$ defined by $\nabla_{\gamma'}\gamma'$. One can show a geodesic is a curve that is solution to the equation of geodesics
$$\nabla_{\gamma'}\gamma'=0$$
Once again, if $M\subset \mathbb{R}$, a curve $\gamma : I \to M$ is a geodesic if and only if its acceleration is orthogonal to $M$, if and only if $M$ is a solution of the equation of geodesics. Thus, saying $\nabla_{\gamma'}\gamma'=0$ is a generalisation of saying the acceleration of $\gamma$ is orthogonal to $M$: if there was an ambiant space, it would be.
If for example $M = \mathbb{S}^n \subset \mathbb{R}^{n+1}$, a geodesic is a great circle of the sphere. It can be parametrized by
$$\gamma(t) = \cos(t\|v\|)p + \sin(t\|v\|)\frac{v}{\|v\|}$$
where $p\in \mathbb{S}^n$ and $v \in T_p\mathbb{S}^n = p^{\perp}$. The usual differentiation of $\gamma$ in the ambiant space shows that
$$ \gamma''(t) = -\|v\|^2\cos(t\|v\|)p -\|v\|^2\sin(t\|v\|)\frac{v}{\|v\|}=-\|v\|^2\gamma(t) $$
which is colinear to $\gamma(t)$ and thus, orthogonal to $T_{\gamma(t)}\mathbb{S}^n$. Its orthogonal projection onto $T_{\gamma(t)}\mathbb{S}^n$ is then $0$.
Best Answer
Let $\nabla$ be the Levi-Civita connection of $(M,g)$ and $c$ be a geodesic of $(M,g)$, then by definition, one has: $$\overline{\nabla}_{\dot{c}}\dot{c}=0,$$ where $\overline{\nabla}$ stands for $c^*\nabla$, then since $g$ is $\nabla$-parallel, the induced metric $\overline{g}:=c^*g$ is $\overline{\nabla}$-parallel, so that: $$\frac{\mathrm{d}}{\mathrm{d}t}\overline{g}(\dot{c},\dot{c})=2\overline{g}(\nabla_{\dot{c}}\dot{c},\dot{c})=0,$$ therefore $c$ is parametrized at constant speed that is proportionally to the arc length. I want to insist on this last point, a geodesic is not necessarily parametrized at speed $1$. This little discussion established that geodesic are not purely geometric objects.
The geodesic equation stated by Do Carmo is simply the coordinate-dependent version of $\overline{\nabla}_{\dot{c}}\dot{c}=0$. Indeed, by the very definition of the Christoffel $(2,1)$-tensor, one has $\nabla=\nabla^0+\Gamma$, where $\nabla^0$ is the flat connection of the given chart of $M$. Therefore, as we saw, being a solution of the geodesic equation implies being parametrized proportionally to the arc length, so no need to assume that the curve satisfies this condition.
Feel free to let me know if I totally missed the point of your question.