Well, you only need to know some properties of the tensor product:
$$
(A_1 \otimes B_1)\cdot(A_2 \otimes B_2) = (A_1 \cdot A_2)\otimes(B_1 \cdot B_2),
$$
where I left explicit that $a \cdot b$ is the normal matrix multiplication between $a$ and $b$. It is understood that $A_1$ and $A_2$ live in the same space, i.e. it could be $A_1,A_2 \in \mathbb{C_{2 \times 2}}$. (Same applies for $B_1$ and $B_2$, i.e. they belong to the same space). Note $A_1$ and $B_1$ may not belong to the same space.
Lets take the first term:
$$
b_1^\dagger b_2 = (I \otimes I \otimes \sigma^+)(I \otimes \sigma^- \otimes \sigma^z)=(I \otimes \sigma^-\otimes \sigma^+ \sigma^z) \equiv -(I \otimes \sigma^-\otimes \sigma^+),
$$
cause
$$
\sigma^+ \sigma^z = -\sigma^+.
$$
Do that for all terms and use the distributive property
$$
A_1\otimes(B_1 + B_2) = A_1 \otimes B_1 + A_1 \otimes B_2, \tag{*}
$$
remembering that
$$
\sigma^+ = (\sigma^x+i\sigma^y)/2
$$
so $b_1^\dagger b_2$ really is
$$
b_1^\dagger b_2 = -(I \otimes \sigma^-\otimes \sigma^+) = - (I \otimes (\sigma^x-i\sigma^y)/2\otimes (\sigma^x+i\sigma^y)/2),
$$
and I'll let you expand that one out. Spoiler alert:
$$
b_1^\dagger b_2+b_2^\dagger b_1 = I\otimes \sigma^x \otimes \sigma^x + I \otimes \sigma^y \otimes \sigma^y.
$$
