Bound on $\|f(x)\|$ from a bound on $\|f'(x)\|$. 
Let $B$ be a Banach space, let $f : [0,+∞) → B$ be a continuous
  function differentiable on $(0,+\infty)$, and let $g : [0,+∞) → [0,+∞)$ be a nondecreasing function differentiable on $(0,+∞)$, such
  that:
  
  
*
  
*$\|f(0)\| ≤ g(0)$,
  
*$∀ t ∈ (0,+\infty), \|f'(t)\| ≤ g'(t)$.
  
  
  Can we conclude that $∀ t ∈ [0,+∞), \|f(t)\| ≤ g(t)$?

We can use the mean inequality to show the weaker statement that if $\|f'(t)\| = o(t^n)$ as $t→0$ and $f(0)=0$, then $\|f(t)\| = o(t^{n+1})$. If we assume $f$ and $g$ to be $\mathcal{C}^1$, then we can use the bound $\|f(t)\| = \left\|f(0) + ∫_0^t f'(t) \mathrm{d}t\right\| ≤ \|f(0)\| + ∫_0^t \left\| f'(t) \right\| \mathrm{d}t$ to conclude.
I'm also interested in proving the following statement if it is true (and also in similar ones if it's not too broad of a question).

Let $G : [0,+∞) × (0,+∞) → [0,+∞)$ be a locally Lipschitz function in
  its first variable. Suppose $f : [0,+∞) → B$ and $g : [0,+∞) → [0,+∞)$
  are such that:
  
  
*
  
*$f$ and $g$ are continuous on $[0,+∞)$ and differentiable on $(0,+∞)$,
  
*$\|f(0)\| ≤ g(0)$,
  
*$\|f'(t)\| ≤ G(\|f(t)\|, t)$,
  
*$g'(t) = G(g(t),t)$ (or $≥$).
  
  
  Then $∀ t ∈ [0,+∞), \|f(t)\| ≤ g(t)$.

Here is a related question : Comparison theorem for ODE. The main problem here is that we cannot differentiate $\|f(t)\| - g(t)$. Is there a weaker notion than the derivative that can help?
Let us suppose we can differentiate $h(t) := \|f(t)\| - g(t)$ and that $h'(t) ≤ \|f'(t)\| - g'(t)$. Let $H(y,t) := G(y+g(t),t) - G(g(t),t)$. Then $H$ is locally Lipschitz in $y$ and $h'(t) ≤ H(h(t),t)$. Moreover $H(0,t)=0$ for all $t$ and $h(0)≤0$. Let $M$ be the Lipschitz constant of $H$ near $0$. Suppose that $h(t)>0$ for some $t>0$ and suppose WLOG that $h$ is positive on $(0,t]$ (otherwise, translate the function). Let $t_0>0$ be small enough such that $t_0 M < 1$ and $h(t_0)≥h(t)$ for all $0≤t≤t_0$ (this is possible by continuity of $h$). Then for all $0≤t≤t_0$, we have $h'(t) ≤ M h(t) ≤ M h(t_0)$ by Lipschitz continuity of $H$. So by the mean inequality, $h(t_0) ≤ M h(t_0) t_0 < h(t_0)$, a contradiction. So we must have $h(t) ≤ 0$ for all $t$. (Note that this can be used as a proof of uniqueness of the solution to a Cauchy problem.)
 A: 
Can we conclude that $∀ t ∈ [0,+∞), \|f(t)\| ≤ g(t)$?

Yes. Consider the function 
$$
 h: [0, +\infty) \to \Bbb R \, , \quad h(t) = \Vert f(t) \Vert - g(t) \, . 
$$
$h$ is continuous with $h(0) \le 0$, and we want to show that $h(t) \le h(0)$ everywhere. $h$ is not necessarily differentiable, but we can mimic the proof of the mean value theorem.
So assume that $h(t_1) > h(0)$ for some $t_1 > 0$. The (continuous) function
$$
 \varphi: [0, t_1] \to \Bbb R \, , \quad \varphi(t) = h(t) - \frac{h(t_1)-h(0)}{t_1} t
$$
satisfies $\varphi(0) = \varphi(t_1) = h(0)$, and assumes its maximum at some point $t_0 \in (0, t_1]$. Then for $0 < t < t_0$:
$$
 \varphi(t) \le \varphi(t_0)
$$ and therefore
$$
 0 < \frac{h(t_1)-h(0)}{t_1} \le \frac{h(t_0)-h(t)}{t_0-t} 
 = \frac{\Vert f(t_0) \Vert - g(t_0) - \Vert f(t)\Vert + g(t)}{t_0-t} \\
 \le \frac{\Vert f(t_0) - f(t) \Vert - (g(t_0) - g(t))}{t_0-t} \\
 = \left\Vert \frac{f(t_0) - f(t)}{t_0-t} \right\Vert -  \frac{g(t_0) - g(t)}{t_0-t} \, ,
$$ 
using the triangle inequality for $\Vert \cdot \Vert$. Now for $t \to t_0$ the right-hand side has the limit
$$
 \Vert f'(t_0) \Vert - g'(t_0) \le 0
$$
and that yields the contradiction.
The essential idea is to use that a “Dini derivative” of $t \to \Vert f(t) \Vert$ is bounded by $g'(t)$, and that this is sufficient to show that $\Vert f(t) \Vert - g(t)$ is decreasing. See also A function with a non-negative upper derivative must be increasing?.
Perhaps this also gives some ideas how to tackle the second problem.
