Let $M$ be a geodesically complete manifold and $m\in M$. The exponential map at $m$ is $$ \exp_m: T_mM\rightarrow M \\ v \mapsto \gamma(1) $$ where $\gamma$ is the geodesic $\gamma:\mathbb{R}\rightarrow M$ such that $\gamma(0)=m$ and $\dot{\gamma}(0)=v$.
Now, I read that $$ \int_0^1 \parallel \dot{\gamma}(t) \parallel dt = \int_0^1 \parallel v \parallel dt = \parallel v \,\,\parallel $$
I do not know why $\dot{\gamma}(t)=v$ for all $t$. It seems that I am missing something quite fundamental here.
The claim is not that $\dot{\gamma}(t) = v$ for all $t$; it's that these two vectors have the same length.
The reason for this equality is that geodesics are constant-speed curves, for otherwise they'd have acceleration in the tangential direction, which is one of the key things that the definition disallows. And since $\|\dot{\gamma}(0) \| = \| v\|$, we get that $\| \dot{\gamma}(t) \| = \| v\|$ for all $t$.