(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.