Let $g:(S^n, *) \to (X, x)$ be a based map between based topological spaces. Suppose that $g$ is null homotopic in $Top$ (therefore homotopic to a constant map).
I want to prove that in this case $g$ is also null homotopic in $Top_{\bullet}$.Therefore the homotopy map can be choosed such that it conserves the basepoint.
My proof: Let $H: S^n \times I \to X$ homotopy between $g(z) = H(z, 0)$ and constant $c = H(z, 1)$. Since $g$ based we have $H(*,0) = g(*) =x$.
Consider $w:= H \vert {\{*\} \times I}$. Therefore $w(0)=x$.
We define a new homotopy $\widetilde{H}: S^n \times I \to X$ by
$\begin{equation} (z,t) \to \begin{cases} H(z,2t) & \text{for } 0 \le t \le 1/2 \\ w(2-2t) & \text{for } 1/2 \le t \le 1 \end{cases} \end{equation}$
This defines a homotopy from $g$ to constant map to $x$.
But what we really need is a homotopy $\bar{H}: S^n \times I \to X$ with $\bar{H}(*, t) =x$ for all $t \in I$.
By quotient property $\widetilde{H}$ corresponds to a map $m:CS^n \to X$ where $CS^n = S^n \times I /S^n \times \{1\}$. What we really want is that $m$ factorises through $K := S^n \times I /S^n \times \{1\} \cup \{*\} \times I$.
The next step is to call in mind that can inclusion $\{*\} \to S^n$ is a cofibration, therefore also $\{*\} \times I \to S^n \times I $ (see retract characterisation of cofibration).
First question: Is then also $\{*\} \times I \to S^n \times I /S^n \times \{1\} = CS^n$ a cofibration? Why?
If that is the case then one knows that since $\{*\} \times I $ is contractable we have $CS^n = S^n \times I /S^n \times \{1\} \cong S^n \times I /S^n \times \{1\} \cup \{*\} \times I = K$
So we get indeed a map $ K \to X$ with $ K \cong CS^n$.
Now the second problem: the isomorphism $ K \cong CS^n$ isn't canonical so I don't know how extract from $K$ the desired based homotopy $\bar{H}$ between $g$ and $x$.
Your approach is not bad, but is too complicated. Since $g$ is null homotopic in $Top$, it has a continuous extension $G : D^{n+1} \to X$. Define
$$R : S^n \times I \to D^{n+1}, R(y,t) = t \ast + (1-t)y ,$$
$$H = G \circ R : S^n \times I \to X.$$
Then $H(y,0) = G(R(y,0)) = G(y) = g(y)$, $H(y,1) = G(R(y,1)) = G(\ast) = x$, $H(\ast,t) = G(R(\ast,t)) = G(\ast) = x$.
By the way, $\{*\} \times I \to S^n \times I /S^n \times \{1\} = CS^n$ is a cofibration. You can see this directly by the retract characterization (perhaps somewhat tedious) or by observing that $\{*\} \times I$ is mapped to a radial line segment $L$ in $CS^n \approx D^{n+1}$. But you can give $D^{n+1}$ a CW-structure such that $L$ is s a subcomplex, thus $L \hookrightarrow D^{n+1}$ is a cofibration.