Jacobi fields are linearly independent if and only if... 
If the dimension of the Riemannian manifold $M$ is $n$, there exist exactly $n$ linearly independent Jacobi fields along the geodesic $\gamma : [0,a] \to M$, which are zero at $\gamma(0)$. This follows from the fact, easily checked, that the Jacobi fields $J_1,\ldots,J_k$ with $J_i(0)=0$ are linearly independent if and only if $J_1'(0),\ldots,J_k'(0)$ are linearly independent.

(Riemannian geometry, Manfredo do Carmo, page 116)
I tried to "prove" first the reverse direction:
If $J_1'(0),\ldots,J_k'(0)$ are linearly independent, then $\sum_i c_i J_i'(0)=0 \iff c_i=0$. Applying the limit definition and $J_i(0)=0$, we have that $\lim_{t \to 0^+} \frac{\sum_i c_iJ_i(t)}{t}=0$. Then I used the epsilon-delta definition of the limit: $\forall \epsilon > 0, \exists \delta > 0$ such that $|\frac{\sum_i c_i J_i(t)}t-0|<\epsilon$. Well, $t$ is defined only on $[0,a]$, so $\frac 1t \ge \frac 1a$, and so $\frac{|\sum_i c_i J_i(t)|}a\le|\frac{\sum_i c_i J_i(t)}t-0|<\epsilon$, and so $|\sum_i c_i J_i(t)| < \epsilon a$. Since the inequality we just deduced holds true for all $\epsilon > 0$, we must have that $\sum_i c_i J_i(t) = 0$ if $0 < t < \delta$. Since $c_i=0$, it follows that $\sum_i c_i J_i(t) = 0$ for all $0 \le t \le a$.

I don't feel good about this alleged proof; I know there are some statements that don't follow (although they seemed like they would to me). Furthermore, how would I also go about proving the other direction? Please do let me know if there is a better proof that I should follow.

Here is an alternative proof. We know the explicit form of a Jacobi field $J$ with $J(0)=0$. See Corollary 2.5 of Chapter 5 of Do carmo. Now use this explicit formula with the fact that derivative is a linear map. At $t=0$ they are linearly independent by definition. At other points $t\neq 0$ therefore we can use formula.
