Reaction with tin(IV) iodide and hydrogen peroxide
Note: this is a preliminary description, which I think answers the question, but disagrees in one detail (see asterisk).

Tin(IV) iodide is completely hydrolyzed by water [1, p. 120] to colorless hexahydroxostannic acid:

$$\ce{SnI4 + 6 H2O &lt;=&gt; H2[Sn(OH)6] + 4 HI}$$

Hydrogen peroxide can not oxidize $\ce{Sn^4+}$ further (in fact, hexahydroxostannate(IV) is used as a stabiliser for hydrogen peroxide), and is likely going to oxidize hydroiodic acid to iodine (which attributes to the brown color) instead:

$$\ce{H2O2 + 2 HI -&gt; I2 + 2 H2O}$$

Hexahydroxostannic acid only exists in solution and tends to form cloudy/milky gel over time [2, p. 1021]; this can be simplified as formation of colloidal particles of tin(IV) oxide [hydrates]:

$$\ce{H2[Sn(OH)6] &lt;=&gt; SnO2 · n H2O + $(4-n)$ H2O}$$

When petroleum ether is added, iodine is extracted to organic phase (on top due to lower density of the ether). Tin(IV) oxide sol remains in aqueous phase and attributes to its slightly pink color due to Mie scattering of colloidal solution.*

* Phase order doesn't match OP's description which suggests opposite coloring of the phases (additional clarification on the type of petroleum spirit or a photo of the binary phase mixture would help).

References


Inorganic Syntheses; Bailar, J. C., Ed.; Inorganic Syntheses; John Wiley &amp; Sons, Inc.: Hoboken, NJ, USA, 1953; Vol. 4. https://doi.org/10.1002/9780470132357.  
Holleman, A. F.; Wiberg, E.; Wiberg, N. Lehrbuch der anorganischen Chemie, 102nd ed.; Fischer, G., Ed.; Walter de Gruyter: Berlin New York, 2007. ISBN 978-3-11-017770-1. (in German)