I want to get a more geometrical sense in this proof of Gauß's Lemma, which states:
Lemma: The geodesics trough a point $q$ in a Riemannian Manifold in $\epsilon-$ball around $p$ are perpendicular to he hypersurfaces $$\{\exp_q(v): \Vert v \Vert=\text{constant}<\epsilon \}$$ for sufficiently small $\epsilon$
Here is a proof, which I want to better understand:
Let $v:\mathbb{R}\to T_qM$ be a smooth curve with $\Vert v(t) \Vert=\kappa<\epsilon$ for a constant $\kappa$ for all $t$, now define $$\alpha(u,t)=\exp_q(u \cdot v(t)),\quad \quad -1<u<1 $$ By a brute force calculation (which is messy, but straight-forward), we get $$\frac{\partial}{\partial u}\Bigr\langle \frac{\partial\alpha}{\partial u},\frac{\partial \alpha}{\partial t} \Bigr\rangle=0 \quad \text{and } \quad \frac{\partial}{\partial t}\Bigr\langle \frac{\partial\alpha}{\partial u},\frac{\partial \alpha}{\partial t} \Bigr\rangle=0$$ Note that $\alpha(0,t)=\exp_q(0)=q$, thus $\frac{\partial \alpha}{\partial t}\bigr|_{(0,t)}=0$. If follows that $$\Bigr\langle \frac{\partial\alpha}{\partial u},\frac{\partial \alpha}{\partial t} \Bigr\rangle=0 \quad \square $$
Okay, this is simple, the only detail that remains unsettled is the intuition for our choice of $\alpha$. What I mean by this, is that I do not quite understand why $\frac{\partial \alpha}{\partial u}$ and $\frac{\partial \alpha}{\partial t}$ are the "geodesical" and "tangent" components of the variation $\alpha$ of geodesics through $q$ respectively (that is if I even got that right). To be more precise, what does it mean to go through $\alpha$ by $u$ or by $t$?
This is an easy exercise in geometric intuition. You should try to draw a picture in the $2$-d case.
First you should note that $v$ is a curve in $T_qM$, not in $M$ (as you have written).
Since it's position vector has constant length, it is, actually, a curve in the sphere of radius $\kappa$ in $T_q M$.
$u\mapsto uv(t)$, on the other hand, is (note $t$ is fixed) a ray emanating from the origin of $T_qM$. So the family $(u,t) \mapsto u v(t)$ is a trivial variation of the ray for $v(0)$.
The exponential map $\exp_q(uv (t)$ is then mapping the curves of this variation to geodesics starting in $q$ with initial tangent vector $v(t)$, so this is a variation of the geodesic $u\mapsto \exp_q(uv(0))$.
Now since $|v(t)|$ is constant, the derivative of $v(t)$ is tangent to the sphere with radius $\kappa$ in $T_q M$ (so, strictly speaking, it is an element of $T_{v(t)}T_qM$, but this is identified with $T_qM$ in the obvious way. This is possible since $T_qM $ is a vector space). So this is orthogonal to the geodesic ray for $t=0$ (both in $T_qM$ and then in $M$, after applying the exponential map).