Not an answer but definitely too long for a comment. I'll edit this whenever something new comes up. Any hint, help or comment is appreciated.
Question i)
We consider $(M,g)$ a Riemannian manifold.
Define $\gamma(t) = \exp_p(tv)$, (suppose $\|v\|=1$ for convinience) and $Y(t) = \mathrm{d}\exp_p(tv)\cdot tw$. It is well known that $Y$ is a Jacobi field along $\gamma$, that is:
$$
Y'' =-R_{\gamma'}Y \left(:= -R(\gamma',Y)\gamma'\right),
$$
and that it is the unique one satisfying $Y(0) = 0, Y'(0) = w$.
You may have another convention for $R$, but what is important is that $g\left(R_{\gamma'}X,X\right) = \sec(\gamma',X)\|X\|^2$.
Finally, we suppose that $M$ is Hadamard, (and it follows that $Y(t) \neq 0$ if $t \neq 0$) and that $\|Y(1)\| = \|w\|$.
Step 1. Let $f(t) = \|Y(t)\|$ for $t\in [0,1]$. Then $f$ is a convex function. Indeed, $f$ is smooth whenever $Y \neq 0$, thus is smooth on $]0,1]$. Moreover, on $]0,1]$:
\begin{align}
f' &= \frac{g(Y',Y)}{\|Y\|} \\
f'' &= \frac{\left(g(Y'',Y) + \|Y'\|^2 \right)\|Y\| - g(Y',Y) \frac{g(Y',Y)}{\|Y\|}}{\|Y\|^2} \\
&= \frac{-\sec(\gamma',Y) \|Y\|^4 + \|Y'\|^2\|Y\|^2 - g(Y',Y)^2}{\|Y\|^3}\\
&= -\sec(\gamma',Y)\|Y\| + \frac{\|Y'\|^2\|Y\|^2 - g(Y',Y)^2}{\|Y\|^3}.
\end{align}
From the Cauchy-Schwarz inequality, the last term is $\geqslant 0$, and from the Hadamard hypothesis, $\sec(\gamma',Y) \leqslant 0$, hence:
$$f'' \geqslant 0,$$
and $f$ is convex.
Step 2. We have an upper bound on $f$. It is clear that $f(0) = 0$ and $f(1) = \|w\|$. By convexity:
$$
\forall t \in [0,1], f(t) \leqslant (1-t)f(0) + tf(1) = t\|w\|,
$$
hence, $f(t) \leqslant t \|w\|$.
Step 3. This latter inequality is an equality. To see this, define:
$$
h(t) = \begin{cases} \frac{f(t)}{t\|w\|} & \text{if } t>0 \\ 1 & \text{if } t =0 \end{cases}.
$$
Then $h$ is continuous (because $Y(t) \sim_0 tY'(0)$), smooth on $]0,1]$, and:
$$
h'(t) = \frac{tf'(t) - f(t)}{t^2\|w\|}.
$$
Write $tf'(t) - f(t) = tf'(t) - f(t) - (0\times f'(0) - f(0))= \int_0^t \left(sf'(s) - f(s)\right)'\mathrm{d}s = \int_0^t sf''(s)\mathrm{d}s$. It follows, from the convexity of $f$, that $h' \geqslant 0$. Hence, $h$ is non-decreasing and:
$$
\forall t \in [0,1], h(t) \geqslant h(0) = 1
$$
that is:
$$
\forall t \in [0,1], f(t) \geqslant t\|w\|.
$$
Finally, $f(t) = t\|w\|$ for all $t \in [0,1]$.
Step 4. $\forall t \in ]0,1]$, $\sec(\gamma'(t),Y(t)) = 0$. Indeed, $f$ is linear, hence $f''=0$. it follows that:
$$
0 = f'' = -\sec(\gamma',Y)\|Y\| + \frac{\|Y'\|^2\|Y\|^2 - g(Y',Y)^2}{\|Y\|^3}\geqslant 0
$$
and as all terms are $\geqslant 0$ on the RHS, we can conclude that $\sec(\gamma',Y) = 0$.
Note that we are in presence of the equality case in the Cauchy-Schwarz inequality, and we can therefore claim that $Y(t)$ and $Y'(t)$ are linearly dependant.
From now, I cannot conclude that $Y$ is colinear to $E_w(t)$, the parallel transport of $w$ along $\gamma$ ; it seems natural in this case, but I guess I'm missing the trick.
Update 1.
We have already shown that $Y'$ and $Y$ are colinear. Hence, there exists $\alpha$, a function defined on $(0,1]$, with $Y' = \alpha Y$. Differentiating this equality gives:
$$
Y'' = (\alpha' + \alpha^2)Y,
$$
and taking the scalar product with $Y$ gives, recalling that $\sec(\gamma',Y) = 0$:
$$
\alpha' + \alpha^2 = 0
$$
It follows that $\alpha(t) = \frac{1}{t - c_0}$ for a constant $c_0$. The initial data give $c_0=0$, i.e $\alpha(t) = \frac{1}{t}$.
Summary of what we have shown for now.
- the function $t \in [0,1] \mapsto \|Y(t)\|$ is linear
- $\forall t\in [0,1], ~\sec(\gamma'(t),Y(t)) = 0$
- $\forall t \in [0,1]~, tY'(t) = Y(t)$
Best Answer
Regarding your comment above: yes, if $M$ is complete and connected with non-postiive curvature, then the universal cover $\widetilde{M}$ of $M$ inherits a complete connected non-positive curvature metric (just pull-back the Riemannian metric from $M$), and so satisfies the conditions (and so also the conclusions!) of the Cartan--Hadamard theorem. One then finds that $\widetilde{M}$ is homeomorphic to the tangent space of any point $\widetilde{m}$ of $\widetilde{M}$ (via $\exp_{\widetilde{m}}$. If $m$ is the image of $\widetilde{m}$ in $M$, then the tangent space to $\widetilde{m}$ is naturally identified with the tangent space to $m$, and the exponential map $\exp_m$ is naturally identified with the composite of $\exp_{\widetilde{m}}$ and the projection $\widetilde{M} \to M$.
Thus indeed, one finds that $\exp_m$ is a covering map.
This circle of ideas is frequently applied, e.g. in the context of hyperbolic manifolds. If $M$ is a compact connected hyperbolic manifold, then one finds by Cartan--Hadamard that $\widetilde{M}$ is isometric to hyperbolic space of the appropriate dimension, and so $M$ is isometric to a quotient of hyperbolic space by a discrete cocompact group of isometries. (This is why hyperbolic manifold theory interacts so tightly with certain parts of group theory.)
Also, one concludes that if $M$ is compact and connected (and positive dimensional --- i.e. not a single point!) with non-positive curvature then $\pi_1(M)$ is infinite (because, writing $M$ as a quotient of $T_m M$ via $\exp_m$, we see that to get something compact, we have to quotient out by an infinite group of diffeos).