(This is a long partial answer which at least gives a concrete example showing that two implications which you probably don't expect to be true are in fact not true)
To get a clear language going, let's be very explicit about naming these axioms.
Let's call by (Waldhausen) the Waldhausen axioms, and if I need to mention them explicitly I'll just write (Waldhausen Cyl 1) and (Waldhausen Cyl 2).
We'll call Gunnar's axioms by (Gunnar) and I'll denote the specific axioms as (Gunnar Cyl 1) and (Gunnar Cyl 2).
For Weibel's axioms, we'll get a little bit more complicated. We'll call the lot by (Weibel), and we'll split the axioms a little more finely into the first three which correspond exactly to what you've listed, so (Weibel Cyl 1), (Weibel Cyl 2), (Weibel Cyl 3). The fourth axiom let's split into:
(Weibel Cyl 4a) : Given a map $(a,b): f \rightarrow f'$ in $\mathcal{C}/\mathcal{C}$, if $a$ and $b$ are cofibrations in $\mathcal{C}$, then so is $T(f) \rightarrow T(f')$.
and
(Weibel Cyl 4b) : Given a map $(a,b): f \rightarrow f'$ in $\mathcal{C}/\mathcal{C}$, if $a$ and $b$ are cofibrations in $\mathcal{C}$, then the map induced by (Weibel Cyl 2) $ A' \coprod_A T(f) \coprod_B B' \rightarrow T(f')$.
Okay, so with this language in place it follows quickly from the axioms of a Waldhausen category that (Gunnar) $\Rightarrow$ (Weibel), and also that (Waldhausen) $\Rightarrow$ (Weibel Cyl 1) + (Weibel Cyl 2) + (Weibel Cyl 3) + (Weibel Cyl 4a).
At a glance, it feels that it should be true that (Waldhausen) $\Rightarrow$ (Gunnar), but I have a toy example that I thought contradicts this, but does not. It does contradict (Weibel) $\Rightarrow$ (Waldhausen) and (Weibel) $\Rightarrow$ (Gunnar) which I certainly expected anyway.
Let $\mathcal{C}$ be the category of pointed finite sets, with $\mathrm{cof}\,\mathcal{C} = \mathrm{w}\,\mathcal{C}$ to be the injections. This gives a non-saturated Waldhausen category, and we can give it a cylinder functor which for $f: A \rightarrow B$ is defined to be $A \times B$, with $j_1 = \mathrm{id}_A \times f$, $j_2 = \{a\} \times \mathrm{id}_B$, and $p = \pi_2$. (You have to be slightly more careful here to get set-theoretic equality for (Waldhausen Cyl 2) but it's fixable). For a morphism of arrows $(g,h) : f \rightarrow f'$, define $T((g,h))$ to simply be $g \times h$.
To see that we get most of (Waldhausen Cyl 1): notice $A \vee B$ always injects into $A \times B = T(f)$, so that we always take cofibrations to cofibrations and weak equivalences to weak equivalences. Clearly $pt \rightarrow pt \mapsto pt \rightarrow pt$ from $\mathrm{Ar}\, \mathcal{C}$ to $\mathrm{F}_1 \, \mathcal{C}$ so that it takes zero to zero. We just need to check that it preserves pushout's along cofibrations, but it does not. Thus this is not a Waldhausen cylinder functor.
It can't be a Gunnar cylinder functor either for cardinality reasons. Given a diagram $B \leftarrow A \rightarrow C$ the pushout $D$ is a quotient of $B \vee C$ and so $|D| \leq |B|+|C|$, but $T(f)$ grows like $|B|\cdot |C|$, and so picking relatively large $X',Y'$ means that $T(f')$ can be much larger than $|T(f)|+|X'\vee Y'| < |T(f)|+|X'|+|Y'|$.
Now you can quickly check that we satisfy (Weibel), and you get the proposition.
I suspect that in general (Waldhausen) does not imply either (Weibel Cyl 4b) or (Gunnar), and that (Gunnar) doesn't imply (Waldhausen) but I suspect you'll need even stranger examples. In any case, all three do cover the important examples that we care about the most.