How do I identify the compound(s) that formed from a failed synthesis of another compound?
Theoretical part

Chrome alum $\ce{KCr(SO4)2 * 12 H2O}$ is synthesized according to the following reaction (average oxidation numbers are shown above the elements undergoing redox reaction):

$$\ce{K2\overset{+6}{Cr}_2O7 + 3 \overset{-2}{C}_2H5OH + 4 H2SO4 →[15-40~°C] K2SO4 + \overset{+3}{Cr}_2(SO4)3 + 3 \overset{-1}{C}_2H4O + 7 H2O}$$

after joint crystallization of both $\ce{K2SO4}$ and $\ce{Cr2(SO4)3}$ with water forming a complex which is better described as aquacomplex $\ce{[K(H2O)6][Cr(H2O)6](SO4)2}$ with no sulfate(VI) anions in neither of the first coordination spheres.

Note:


It's absolutely crucial to control the temperature during the addition of alcohol to acidified dichromate. If the temperature below 15 °C, the reaction is too slow, but if it's above 40 °C, ligand exchange starts to take place: $\ce{H2O}$ ligands in $\ce{[Cr(H2O)6]^3+}$ are partially substituted by $\ce{SO4^2-}$, which is reflected in color change from blue-violet to green. Also, evaporation of alcohol also happens quicker at elevated temperatures, depleting the reducing agent at higher rate.
Whereas you need stoichiometric amounts of dichromate and sulfuric acid, alcohol is usually taken in excess to ensure complete reduction of $\ce{Cr^6+}$ to $\ce{Cr^3+}$ and to salt out crystals of chrome alum from the solution. 
Working with such small quantities (few mL) of pretty abundant and cheap chemicals seems unreasonable to me as it increases the error of measurement, increases the negative impact of impure chemicals or erroneous concentrations and makes the experiment less demonstrative.
I'd advise to work with larger quantities (e.g. 1-2 g of dichromate) in the future.


Experimental results and discussion


  There were some crystals formed, but the color was orange …


It's extremely likely that the orange crystals are unreacted potassium dichromate(VI), so it looks like the reduction was incomplete.
That's why I doubt that alcohol was ever oxidated further to acetic acid, hence I see the formation of chromium(II) acetate as suggested in the comments as improbable.
Also, chromium(II) acetate is very unstable and would've decomposed under these conditions.

There is also a rudimentary method of checking the oxidized product by its smell.
If it smells like apples, then it's acetaldehyde $\ce{C2H4O}$; if it smells like vinegar, then the oxidation took further and alcohol has been oxidized to acetic acid $\ce{CH3COOH}$.


  … (it should've been purple).


Since you haven't dissolved potassium dichromate in water and just added sulfuric acid to the powder, the concentration of sulfate(VI) was too high.
Plus, you've added all alcohol at once, which might've resulted in temperature spike. Both factors effectively prevent formation of chromium(III) aquacomplex, leaving no chance for alum to crystallize. (see Addendum for the reference)


  From the instruction that we followed, we should heat up the solution, so we did that, and carelessly let the temperature go up past 70 °C. It become greenish in color and doesn't form any crystal after a week.


At this stage you pushed the reduction further and probably evaporated both ethanol and acetaldehyde completely, and the solution was consisting of very concentrated chromium(III) sulfate, still no chance for alum to crystallize due to above mentioned reasons. (see Addendum)


  We resorted to evaporating the liquid part and obtained some leftover greenish yellow powder at the bottom


Green powder is likely a mixture of potassium chromium(III) sulfate $\ce{KCr(SO4)2}$ and also minor amounts of chromium(III) oxide $\ce{Cr2O3}$ formed along the way.
Now yellow powder is with a very high confidence just potassium chromate(VI) $\ce{K2CrO4}$ leftovers from the dichromate(VI) once the acid was used up:

$$\ce{Cr2O7^2- + H2O &lt;=&gt; 2 HCrO4^2- &lt;=&gt; 2 CrO4^2- + 2 H+}$$

One might also claim that it's elemental sulfur formed by reduction of sulfate(VI) after chromium(VI) was completely converted to chromium(III), but I find it rather implausible.   

Addendum

From Ullmann’s Encyclopedia of Industrial Chemistry [1, p. 4441]


  Potassium chromium(III) sulfate [10141-00-1], [10279-63-7], [7788-99-0],
  potassium chrome alum, $\ce{KCr(SO4)2 · 12 H2O}$, $M_\mathrm{r}~499.11$, $\pu{1.813 g/cm3}$, crystallizes in the cubic system forming violet regular octahedra
  which decay in air and melt at 89 °C, the color changing to green; the enthalpy of formation is − 5788 kJ/mol. The solubility in water at 25 °C is 11.1 wt %. The solution is violet when cold, but becomes green above 50 °C, this change being accompanied by a decrease in the molar conductance. For 0.125 M solution at 50 °C, the molar conductance is $\pu{221 Ω−1 cm2 mol−1}$ for the violet form and $\pu{202 Ω−1 cm2 mol−1}$ for the green form. This change is reversible and its rate is increased by acids. The green form always occurs as an amorphous solid and crystals
  are unknown.
  
  […]
  
  For the preparation of potassium chromium alum, a saturated potassium dichromate solution is reduced with sulfur dioxide in the presence of sulfuric acid.
  During the reaction, the temperature must be kept below 40 °C by cooling to prevent the green modification being produced. Apart from sulfur dioxide, such organic compounds as formaldehyde, methanol, or starch are suitable as reducing agents. Crystallization starts after sulfuric acid has been added and the temperature is kept
  further below 40 °C. If the solution is allowed to settle in vats, large crystals are produced, whereas ﬁne ones result if the solution is stirred.


References


Ullmann’s Encyclopedia of Industrial Chemistry, 40 Volume Set, 7th ed.; Wiley-VCH, Ed.; Wiley-VCH: Weinheim, 2011. ISBN 978-3-527-32943-4.