Let $X$ be topological space and $$CX:=(X\times [0;1]) / (X \times \{0\})$$
it's cone. $X$ can via identification $X \cong X \times \{1\}$ be enbedded into $CX$.
I wan't to show explicitely that $i:X \to CX$ is a cofibration by writing concretely the morphism $$m: CX \to Y^I$$
commutating with the morphisms in following diagram:
$$ \require{AMScd} \begin{CD} X @>{H} >> Y^I \\ @VViV @VVev_0V \\ CX @>{f}>> Y \end{CD} $$
Indeed I know that there are a lot of ways to show implicitely that $i$ is a cofibration.
But here I'm interested in the concrete shape of the map $m$.
To set conventions: I will define $CX = X\times I /X\times \{0\}$, I will identify $X \cong X\times \{1\} \subset CX$. We define a retraction $r: CX \times I \to (X\times I) \cup (CX \times \{0\})$ as follows:
$$r(((x,s),t))= \begin{cases} ((x,1), t-2 +2s) & \text{if}\ \ \ 1-\frac{1}{2}t\le s \le 1\\ ((x, \frac{2s}{2-t}),0)& \text{if}\ \ \ 0 \le s \le 1-\frac{1}{2}t \end{cases}$$
It is necessary to check that the function is well defined, though in that case it is clearly continuous. One also needs to check that it is a retraction.
Doing the case of $X = \{*\}$ by drawing a picture of a square and visualizing the retraction onto the left and top edge gives the idea for the formulas.
There is a more sophisticated approach to this type of thing using NDR pairs, and it is fairly easy to prove that $(CX, X)$ is an NDR pair. One advantage is it works out such formulas behind the scenes.
Finally, once you have obtained $r$, how do you obtain $m: CX\to Y^I$? First, note that $H: X\to Y^I$ defines by duality a map $H':X\times I \to Y$ given by $H'(x,t) = H(x)(t)$. The map $f: CX\to Y$ can be viewed as a map $f': CX\times \{0\} \to Y$ in the obvious way, by identification. Thus, we obtain a map $H'\cup f' : (X\times I) \cup (CX \times \{0\}) \to Y$, which agrees on the overlap by the commutativity of your diagram above. Therefore, $H'\cup f'$ is continuous. Now, define $m': CX\times I \to Y$ by $m' = (H'\cup f') \circ r$. The dual map $m: CX \to Y^I$ is the map $m$ you seek. For clarity, the map $m$ is defined by $m((x,s))(t) = m'((x,s),t)$. Hopefully, this clarifies the connection between the retraction and the lifting problem above.