Here you need to think of the pair (surface, metric) as the geometric object, not only of the surface. The upshot here is that the pairs (hyperboloid, euclidean metric) and (hyperboloid, Minkowski metric) are distinct geometric objects. The way to measure lengths and areas is not the same in these two "worlds". The first one has non-constant positive curvature, and the second one has constant negative curvature $-1$.
The reason is that the hyperboloid plays the role of a sphere with respect to the Minkowski metric, since it can be written as the solution set of $\langle p,p\rangle_L=-1$. Compare that with the sphere equation $\langle p,p \rangle_E=1$. If by now you're wondering whether the one-sheeted hyperboloid defined by $\langle p,p\rangle_L=1$ has something special about it, I'll tell you it does: it has constant curvature $1$ when equipped with the Minkowski metric (the one who correctly expresses it as a sphere).
In fact, here's an exercise for you to understand what happens: define $$g((x_1,y_1,z_1),(x_2,y_2,z_2))= 3x_1x_2+5y_1y_2+2z_1z_2$$and show that $M=\{(x,y,z)\in \Bbb R^3\mid 3x^2+5y^2+2z^2=1\}$ has constant curvature when equipped with $g$.
Solution for suggested exercise. Note that $M = \{ p \in \Bbb R^3\mid g(p,p)=1\}$. If $F(p) = g(p,p)$, then $M = F^{-1}(1)$, meaning that the $g$-gradient of $F$ is always $g$-normal to $M$. Since $${\rm d}F_p(v) = 2g(p.v) = g(2p,v),$$we get ${\rm grad}_gF(p)=2p$, and so $N(p)=p$ is an $g$-unit $g$-normal vector along $M$, just like for spheres in Euclidean space. Indeed, $M$ is a $g$-sphere. Note that the curvature of the Levi-Civita connection of $g$ in $\Bbb R^3$ is zero (indeed, all the Christoffel symbols w.r.t. the usual rectangular coordinates are zero -- this actually says that the Levi-Civita connection is the usual one). So, Gauss' Formula says that $$K(v,w) = \frac{g(I\hspace{-.1cm}I(v,v))g(I\hspace{-.1cm}I(w,w)) - g(I\hspace{-.1cm}I(v,w),I\hspace{-.1cm}I(v,w))}{g(v,v)g(v,w)-g(v,w)^2}.$$Just in the same way we do it for the sphere, you can check (or see e.g. page 101 in Chapter 4 in O'Neill's Semi-Riemannian Geometry with Applications to Relativity) that $I\hspace{-.1cm}I_p(v,w) = -g(v,w)N(p)$. Taking $(v,w)$ to be an orthonormal basis for the tangent plane $T_pM$, plugging in above gives $K(p)=1$ as expected.
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
I do not think this is the right idea. On a surface inside $\mathbb R^3$, the line element still is $$ds^2=dx^2+dy^2+dz^2.$$
You see different formulas when you introduce local coordinates on the surface. For example, consider the paraboloid of equation $z=x^2+y^2$. Then, $dz=2xdx+2ydy$ and so $$ dz^2=4x^2dx^2+ 4y^2dy^2 + 4xy(dxdy+dydx).$$ Therefore, the line element on the paraboloid is $$\tag{1} ds^2=(4x^2+1)dx^2 + (4y^2+1)dy^2 + 4xy(dxdy+dydx).$$
In Riemannian geometry, they take this kind of reasoning one step further. A manifold is given as an abstract object, that needs not be included in $\mathbb R^3$ or in $\mathbb R^n$. Which implies that there is no $dx^2+dy^2+dz^2$ anymore. You have to specify an expression such as (1), which in modern terminology takes the name of metric tensor; you do not compute it, like we just did here.