= Solution
Suppose first that $F:D^n\to X$ extends $f$. Regarding $S^{n-1}$ as the boundary of the unit disc,
$$
H(x,t)=F((1-t)x)
$$
is a <homotopy> from $f$ to the constant map with value $F(0)$. Thus $f$ is <null-homotopic map>[null-homotopic]. Conversely, if $H:S^{n-1}\times[0,1]\to X$ is a homotopy from $f$ to a constant, then $H$ is constant on $S^{n-1}\times\{1\}$ and hence descends to the quotient
$$
S^{n-1}\times[0,1]/(S^{n-1}\times\{1\}),
$$
which is the cone on $S^{n-1}$ and is homeomorphic to $D^n$. The descended map extends $f$. This proves the <extension-null-homotopy criterion for a sphere>.
A <universal cover> of $X$ is a covering map $p:\widetilde X\to X$ whose total space is path-connected and simply connected. Let $p_1:(\widetilde X_1,\widetilde x_1)\to(X,x)$ and $p_2:(\widetilde X_2,\widetilde x_2)\to(X,x)$ be two universal covers. The <lifting criterion for a covering space> applies because
$$
(p_1)_*\pi_1(\widetilde X_1)=0
\subseteq(p_2)_*\pi_1(\widetilde X_2)=0,
$$
so $p_1$ has a based lift $h:\widetilde X_1\to\widetilde X_2$ satisfying $p_2h=p_1$. Similarly there is a based lift $k:\widetilde X_2\to\widetilde X_1$ of $p_2$. Both $kh$ and the identity are lifts of $p_1$ that agree at the base point, so uniqueness of lifts gives $kh=1$. Likewise $hk=1$. Hence $h$ is a homeomorphism. This is the <uniqueness of a universal covering space>.
Now let $p:\widetilde X\to X$ be universal with $\widetilde X$ contractible. If an extension $F:|K|\to X$ exists, functoriality of the <fundamental group> gives
$$
f_*=F_*i_*.
$$
Thus the required factorization holds with $\Phi=F_*$.
Conversely, suppose $f_*=\Phi i_*$. We extend over the simplices of $K$ by dimension. The boundary of every 2-simplex is a loop in $|K^1|$ that becomes null-homotopic in $|K|$, so its class lies in $\ker i_*$. The factorization gives
$$
f_*[\partial\sigma]=\Phi i_*[\partial\sigma]=1.
$$
Hence $f|_{\partial\sigma}$ is null-homotopic in $X$ and extends over the disc $|\sigma|$ by the first part. Since distinct 2-simplices meet along the already fixed 1-skeleton, these extensions combine to a map on $|K^2|$.
Inductively suppose the map is defined on $|K^{m-1}|$ for $m\geq3$. For an $m$-simplex $\sigma$, its boundary is $S^{m-1}$, which is simply connected. The boundary map therefore lifts through $p$ by the covering-space lifting criterion. Its lift into the <contractible space> $\widetilde X$ is null-homotopic, so the boundary map itself is null-homotopic and extends over $|\sigma|$. Extending over every $m$-simplex and then every skeleton constructs $F:|K|\to X$. The weak topology on a simplicial complex makes the cellwise map continuous. This proves the <extension criterion into a space with contractible universal cover>.
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