Proof.
We first show that the function $a$ is well-defined: that is, that the map $\partial _{x}: \pi _{1}(S,s) \rightarrow \pi _0( X_ s )$ depends only on the image of $x$ in $\pi _0( X_ s )$. Fix an element $\eta \in \pi _{1}(S,s)$, which we can write as the homotopy class of an edge $v: s \rightarrow s$ in the Kan complex $S$. Let $x$ and $x'$ be vertices belonging to the same connected component of $X_{s}$, so that there exists an edge $u: x' \rightarrow x$ of $X$ satisfying $f(u) = \operatorname{id}_{s}$. We wish to show that $\partial _{x}( \eta ) = \partial _{x'}(\eta )$ in $\pi _0(X_ s)$. Since $f$ is a Kan fibration, we can lift $v$ to an edge $\widetilde{v}: x \rightarrow y$ in $X$. Using the fact that $f$ is a Kan fibration, we can solve the lifting problem
\[ \xymatrix@R =50pt@C=50pt{ \Lambda ^{2}_{1} \ar [r]^-{( \widetilde{v}, \bullet , u)} \ar [d] & X \ar [d]^{f} \\ \Delta ^{2} \ar [r]^-{ s^{1}_0( v)} \ar@ {-->}[ur]^-{\sigma } & S } \]
to obtain a $2$-simplex $\sigma $ of $X$ depicted in the diagram
\[ \xymatrix@R =50pt@C=50pt{ & x \ar [dr]^{ \widetilde{v} } & \\ x' \ar [ur]^{u} \ar [rr]^{ \widetilde{v}' } & & y. } \]
The edges $\widetilde{v}$ and $\widetilde{v}'$ then witness the identities $\partial _{x}(\eta ) = [y] = \partial _{x'}(\eta )$ in $\pi _{0}(X_ s)$.
We now complete the proof by showing that the function $a: \pi _{1}(S,s) \times \pi _0(X_ s) \rightarrow \pi _0( X_ s)$ determines a left action of $\pi _{1}(S,s)$ on $\pi _0(X_ s)$. Note that the identity element of $\pi _{1}(S,s)$ is given by the homotopy class of the degenerate edge $\operatorname{id}_{s}: s \rightarrow s$ of $S$. For each $x \in X_{s}$, we can lift $\operatorname{id}_{s}$ to the edge $\operatorname{id}_{x}: x \rightarrow x$ of $X$, which witnesses the identity $a( [\operatorname{id}_ s], [x] ) = \partial _{x}( [\operatorname{id}_ s]) = [x]$ in $\pi _0(X_ s)$. To complete the argument, it will suffice to show that for every pair of edges $g,g': s \rightarrow s$ of $S$ and every vertex $x \in X_ s$, we have an equality $a( [g'] [g], [x] ) = a( [g'], a( [g], [x] ))$ in $\pi _0(X_ s)$. Since $f$ is a Kan fibration, we can lift $g$ to an edge $\widetilde{g}: x \rightarrow y$ in $X$, and $g'$ to an edge $\widetilde{g}': y \rightarrow z$ in $X$. Since $X$ is a Kan complex, the map $( \widetilde{g}', \bullet , \widetilde{g} ): \Lambda ^2_1 \rightarrow X$ can be completed to a $2$-simplex $\sigma $ of $X$, as depicted in the diagram
\[ \xymatrix@R =50pt@C=50pt{ & y \ar [dr]^{ \widetilde{g}' } & \\ x \ar [ur]^{ \widetilde{g} } \ar [rr]^{ \widetilde{g}''} & & z. } \]
The edges $\widetilde{g}$, $\widetilde{g}'$, and $\widetilde{g}''$ then witness the identities $a( [g], [x] ) = [y]$, $a( [g'], [y]) = [z]$, and $a( [g'] [g], [x]) = [z]$ (respectively), so that we have an equality
\[ a( [g'] [g], [x] ) = [z] = a([g'], [y]) = a( [g]', a( [g], [x] )) \]
as desired.
$\square$