Neighborhood deformation retracts vs cofibrations
I tend to use the subspace $I\times A\cup 0\times X\subseteq I\times X$, as it tends to make the formulas easier to write down with the statement of definition 3.
$4)\Rightarrow 3)$ Taking $f$ as the identity we get a retraction $r:I\times X\rightarrow I\times A\cup0\times X$. Fixing one such we set $u:X\rightarrow I$ to be the map
$$u(x)=\sup_{t\in I}|t-pr_1\circ r(0,x)|,\qquad x\in X.$$
Also let $h:I\times X\rightarrow X$ be the homotopy
$$h(t,x)=pr_2\circ r(t,x),\qquad t\in I,x\in X.$$
Then all required properties are immediate. (Note that I corrected the last part of your statement of definition 3 to match your sources).
$3)\Rightarrow 4)$ We have the maps $u,h$ and need to define a retraction $r$ to the inclusion $A\times I\cup \{0\}\times X\subseteq I\times X$. This is given by
$$r(t,x)=\begin{cases}(0,h(t,x))&t\leq u(x)\\ (t-u(x),h(t,x))& t\geq u(x)\end{cases}$$
You check easily that it is well-defined. Given $f:A\times I\cup0\times X\rightarrow T$ the extension is now $\widetilde f=fr:X\times I\rightarrow T$.
Thus $3)$ and $4)$ are equivalent and imply that the inclusion of the closed subspace $A\subseteq X$ is a cofibration.
$3)\Rightarrow 2)\Rightarrow 1)$ Set $N=u^{-1}([0,1))$ and let $r:N\rightarrow A$ be the map $r(x)=h(u(x),x)$. The required homotopy $ir\simeq id_N$ is $(t,x)\mapsto h((1-t)u(x)+t,x)$.
Now the last implications are not reversible in general. It turns out the presence of the function $u$ is extremely important. If you have $u$, then you can go back, and Aguilar, Gitler and prieto give a proof under the additional assumption that $X$ is perfectly normal (pg. 94 of Algebraic Topology from a Homotopical Viewpoint).
As for your last question, if $(X,A)$ is a closed NDR pair (def. 3), then we have a retraction $r:I\times X\rightarrow I\times A\cup0\times X$, and a homotopy
$$H_s(t,x)=((1-s)t+s pr_1\circ r(t,x),pr_2\circ r(st,x))$$
Thus the inclusion of $I\times A\cup0\times X$ into the cylinder is a strong deformation retraction.