I have the following questions about the Homogenity of geodesic: Let $\gamma:(-\delta,\delta)\to M$, where $t\to\gamma(t,q,v)$ is a geodesic, then $\gamma:\left(-\dfrac{\delta}{a},\dfrac{\delta}{a}\right)\to M$, where $t\to\gamma(t,q,av)$, $a\in\mathbb{R}$ is a geodesic and $\gamma(t,q,av)=\gamma(at,q,v)$.
The proof is the follow, define by $h(t)=\left(-\dfrac{\delta}{a},\dfrac{\delta}{a}\right)\to M$ the curve $t\to h(t)=\gamma(at,q,v)$, this implies that $h'(t)=a\gamma'(at,q,v)$, $h(0)=q$ and $h'(0)=av$. Take the conection $$\dfrac{D}{dt}\left(\dfrac{d\gamma}{dt}\right)=\nabla_{\frac{dh}{dt}}{\dfrac{dh}{dt}}=\nabla_{a\gamma'(at,q,v)}{a\gamma'(at,q,v)}=a^2\nabla_{\gamma'(at,q,v)}{\gamma'(at,q,v)}=0$$ Therefore, $h$ is a geodesic and we're done. But I don't see why $$ \nabla_{a\gamma'(at,q,v)}{a\gamma'(at,q,v)} = a^2\nabla_{\gamma'(at,q,v)}{\gamma'(at,q,v)}, $$ I know that the connection is linear, i.e. $\nabla_{fX+gY}Z=f\nabla_{X}Z+g\nabla_{Y}Z$ is this the reason (for $f$, $g$ in the ring of real functions of class $C^{\infty}$)?
And other questions it's: any differential curve $\gamma(t)$ in $M$ determine a curve $t\to(\gamma(t),\gamma'(t))$ in his tangent bundle $TM$. Thanks!!
As you say, $\nabla_{fX + gY}Z = f\nabla_{X}Z + g\nabla_{Y}Z$ for all smooth $f$ and $g$. Since $a$ is a number, $$ \nabla_{a\gamma'(at,q,v)}{a\gamma'(at,q,v)} = a\nabla_{\gamma'(at,q,v)}{a\gamma'(at,q,v)}. $$ The second factor of $a$ comes out of the covariant derivative because constants factor out of derivatives. (If you prefer, use the Leibniz rule together with the fact that the covariant derivative of a constant vanishes.)
The answer to your second question is "yes".