(N.B. In the below I assume $[a,b] = [0,\infty]$, but the precise values don't matter and appropriate substitutions of $a,b$ into the discussion also gives you the same conclusion.)

Once you have a local existence theorem of the form

For every compact set $K$ and compact subset $K_0 \Subset K$, there exists $T > 0$ such that for every $x_0\in K_0$ there exists a solution to the IVP with initial value $x_0$ defined on $[0,T]$ such that the trajectory remains in $K$ for all $T$

Then upgrading to global existence is simply an issue of proving that any solution cannot escape to infinity in finite time. This is evidenced by the 5th theorem you cited: if you have
$$ |x'| \leq M(t) (1+|x|) $$
then
you have
$$ \frac{|x|'}{1+|x|} \leq M(t) $$
and integrating both sides you find
$$ \ln (1+|x|) \Big|_{t = 0}^{t= T} \leq C $$
using that $M(t)$ is summable.

## Slight refinement.

Let's abstract the argument a bit. Assume you have a bound on $f$, such that you can write

$$ |x'| \leq M(t) g(|x|) $$

for some given function $g:\mathbb{R}\to \mathbb{R}$ that is locally integrable. Denote by $G$ any primitive of the function $1/g$. Then integrating this differential inequality for a solution guarantees

$$ G(|x|) \Big|_{t = 0}^{t = T} \leq C$$

provided that $M(t)$ is summable. And so **as long as $G$ is coercive** (in other words, it is proper) then the same argument as above will guarantee that $|x|$ does not blow-up in finite time and hence you have global existence.

You can therefore do slightly better then $(1+|x|)$, since there are superlinear $g$s for which $1/g$ is not integrable on the entire real line. For example, you may wish to take

$$ g(|x|) = (1 + |x|) \ln (1+|x|) $$

Then you have a primitive

$$ G(s) = \ln \ln (1+s) $$

which is still proper. In fact you can extend this using the usual tower of natural logs.

## Negative examples

But you can certainly not go beyond logarithmic improvements, at least in general.
You can see this easily by considering scalar equations of the form

$$ x' = M(t) (1 + |x|)^{1+\epsilon} $$

where $M$ is a positive summable function.

Note that with this assumption the solution is increasing, and so positive initial data will lead to positive solutions. And so if $x(0) = x_0$ is positive, the solution satisfies

$$ \frac{x'}{(1+x)^{1+\epsilon}} = M(t) $$

which you can integrate to find

$$ \frac{1}{(1+x(T))^\epsilon} = \frac{1}{(1+x_0)^\epsilon} - \epsilon \int_0^T M(t) ~dt $$

And so with $M(t)$ considered fixed, for every $T$ there exists some $R$ such that if $x_0 > R$ then $x$ must blow up prior to $t = T$.

## Small data regime

The previous discussion however hints at a small-data version of the result. Those that for the previous scalar equation, if

$$ (1 + x_0)^\epsilon \leq \frac{1}{\epsilon \int_0^\infty M(t) ~dt} $$

then we guarantee that $(1 + x(t))^{-1}$ is bounded away from infinity and cannot blow-up in finite time. And hence we have a statement of the form:

There exists a compact set $K_0$ with non-empty interior such that for all $x_0 \in K_0$ there exists a (future-in-time) global solution.

In the general form we gave earlier: let's assume without loss of generality that $g(|x|)$ is positive, and hence $G(s)$ is either unbounded (in which case you have that it is proper and global existence for all data) or that $\lim_{s\to \infty} G(s) = G_\infty$ exists.

In the latter case, let $C = \int_0^\infty |M(t)|$. Then provided the set
$$ \{ r : G(r) < G_\infty - C \} $$
is non-empty, any initial data $x_0$ with $|x_0|$ in this set will generate a solution that exists globally.

## Additional Structures

Note also that you may have a situation where you have additional structures. For example, it is possible that the best possible absolute value bound
$$ |f(t,x)| \leq M(t) (1+|x|)^\kappa $$
has $\kappa > 1$, but still you have global existence for all initial data. This would be the case if you happen to have
$$ x\cdot f(t,x) \leq 0 $$
in which case $|x|^2$ is (weakly) a Lyapunov function.