= Solution
Commutation of the two <local flows> means $\phi_t\psi_s=\psi_s\phi_t$ for all sufficiently small independent parameters $s,t$, on their common local domains. Differentiate this identity in $s$ at zero. The <chain rule> gives
$$
d\phi_t|_p(Y_p)=Y_{\phi_t(p)}.
$$
In a <coordinate chart>, differentiating again in $t$ at zero gives $DX_pY_p=DY_pX_p$. But the <Lie bracket of vector fields> has coordinate expression
$$
[X,Y]_p=DY_pX_p-DX_pY_p,
$$
so commuting <local flows> imply $[X,Y]=0$.
For the converse, suppose the <Lie bracket of vector fields> vanishes. Write $u(t)=\phi_t(p)$ in a <coordinate chart>, and put $J(t)=d\phi_t|_p$. Differentiating the <local flow> equation with respect to the initial point gives the variational <ordinary differential equation>
$$
\dot J(t)=DX_{u(t)}J(t),\qquad J(0)=I.
$$
Consequently $W(t)=Y_{u(t)}-J(t)Y_p$ satisfies
$$
\dot W(t)=DY_{u(t)}X_{u(t)}-DX_{u(t)}J(t)Y_p
=DX_{u(t)}W(t),\qquad W(0)=0,
$$
where the zero <Lie bracket of vector fields> was used in the second equality. Uniqueness for this linear <ordinary differential equation> gives $W=0$. This is a local argument and can be continued through successive <coordinate charts>, proving $d\phi_t(Y)=Y\circ\phi_t$ wherever the <local flow> is defined.
For fixed $t$, the curve $s\mapsto\phi_t(\psi_s(p))$ therefore has derivative $Y$ at its current point and starts at $\phi_t(p)$. Uniqueness of the <integral curve of a vector field> $Y$ identifies it with $s\mapsto\psi_s(\phi_t(p))$. Hence \b[$\boxed{\phi_t\psi_s=\psi_s\phi_t\ \Longleftrightarrow\ [X,Y]=0}$ locally]. No <complete vector field> hypothesis is needed.
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