I would like to know how to solve this problem using Burnside's theorem:
A bracelet is to be made by threading four identical red beads and four identical yellow beads onto a hoop. How many different bracelets can be made?
Furthermore how do I know when to apply this theorem? How exactly does it work?
Before we start let us observe that OP asks for a bracelet rather than a necklace which means we have dihedral symmetry rather than just rotational symmetry. We can use either the Burnside lemma or the Polya Enumeration Theorem on this, both require the cycle index $Z(D_8)$ of the dihedral group $D_8$.
Now to compute the cycle index we start with the rotations. Let the slots be numbered zero to seven. There is the identity for a contribution of $a_1^8.$ The rotation that takes $0$ to $1$ makes a contribution of $a_8.$ The one that takes $0$ to $2$ makes a contribution of $a_4^2.$ The one that takes $0$ to $3$ contributes $a_8.$ The one that takes $0$ to $4$ contributes $a_2^4$. From $0$ to $5$ we get $a_8.$ From $0$ to $6$ we get $a_4^2.$ Finally $0$ to $7$ contributes $a_8.$
Next do the reflections. We have four reflections about an axis passing through the centers of two opposite edges, each giving $a_2^4.$ We also have four reflections passing through the centers of opposite vertices for a contribution of $a_1^2 a_2^3.$
Collect the reflections and the rotations to obtain the cycle index
$$Z(D_8) = \frac{1}{16} (a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8 + 4 a_2^4 + 4 a_1^2 a_2^3) \\ = \frac{1}{16} (a_1^8 + 5 a_2^4 + 2 a_4^2 + 4 a_8 + 4 a_1^2 a_2^3).$$
Now to apply Burnside we must compute the number of colorings containing four red beads and four yellow beads fixed by each type of permutation. For $a_1^8$ we must choose four red singletons, giving ${8\choose 4}.$ For $a_2^4$ we must choose two of the four cycles to be red, for a contribution of $5\times {4\choose 2}.$ For $a_4^2$ we have two possiblities giving $2\times 2.$ No contribution from $a_8.$ Finally for $a_1^2 a_2^3$ the singletons are either both red or both yellow, giving $2\times 4\times {3\choose 2}.$ We get for the answer
$$\frac{1}{16} \left({8\choose 4} + 5\times {4\choose 2} + 2\times 2 + 2\times 4\times {3\choose 2}\right) = 8.$$
Here we used the fact that to be fixed by a given permutation an assignment must be constant on the cycles.
The same result can be obtained from the Polya Enumeration Theorem (PET). We have
$$Z(D_8)(R+Y) = 1/16\, \left( R+Y \right) ^{8}+{\frac {5\, \left( {R}^{2}+{Y}^{ 2} \right) ^{4}}{16}}\\+1/8\, \left( {R}^{4}+{Y}^{4} \right) ^{2} +1/4\,{R}^{8}+1/4\,{Y}^{8}+1/4\, \left( R+Y \right) ^{2} \left( {R}^{2}+{Y}^{2} \right) ^{3} \\= {R}^{8}+{R}^{7}Y+4\,{R}^{6}{Y}^{2}+5\,{R}^{5}{Y}^{3}+8\,{R}^{4} {Y}^{4}\\+5\,{R}^{3}{Y}^{5}+4\,{R}^{2}{Y}^{6}+R{Y}^{7}+{Y}^{8}.$$
We see that $$[R^4 Y^4] Z(D_8)(R+Y) = 8.$$ Here we have used the standard substitution $a_d = R^d + Y^d.$
Remark. When manual computation becomes too cumbersome there is the formula
$$Z(C_n) = \frac{1}{n}\sum_{d|n} \varphi(d) a_d^{n/d}$$
for the cycle index of the cyclic group $Z(C_n)$ of order $n$ which leaves the reflections, which are easy.