Prove: If the map $f_n$ is defined on the $n$-skeleton $X_n$ and you want to define it on an $n+1$-cell with attaching map $ϕ:S_n→X_n$, then you can do so if and only if $f_nϕ$ is nullhomotopic.
I have a good geometric intuition on what is going on: I define the map over the new cell while contracting simultaneously the boundary of $D^{n+2}$, the image of the characteristic map and the composition of this map with $f_n$. By the way I can't find a clear proof of this. As far as I know Hatcher talks about it two times in the book Algebraic Topology but never proving it. I also know there is one quite relevant post here "find the map which induces the homorphism between fundamental groups" but it's too specific. I can't find anything on the web.
Thanks!
If the composition $$f\circ \phi:\mathbb{S}^n\to X_n\to Y$$ is nulhomotopic, then it extends to the disk, i.e., there exists $\bar f:D^{n+1}\to Y$ such that if $i:\mathbb{S}^n\to D^{n+1}$ denotes the inclusion, the obvious diagram commutes: $f\phi = \bar fi$.
Now recall that $X_{n+1}=X_n \cup_\phi D^{n+1}$ is the pushout of the diagram
\begin{array} A\mathbb{S}^{n} & \stackrel{\phi}{\longrightarrow} & X_n \\ \downarrow{i} & & \downarrow{v} \\ D^{n+1} & \stackrel{u}{\longrightarrow} & X_{n+1} \end{array}
The universal property of the pushout implies (as $f\phi=\bar fi$) that there exists a unique extension $g:X_{n+1}\to Y$ completing the diagram (that is, extending $f$).
In the other way around, if there exists $g:X_{n+1}\to Y$, then $f\phi=gvi$ writting again the pushout diagram of $X_{n+1}$, and so $f$ is necessarily nulhomotopic because it factors through the contractible space $D^{n+1}$.