Can someone please, carefully, step by step, explain to me how it is possible to just take out this fraction from the covariance operator? I am lost
$$\begin{align}\mathbb E_t\left[\widetilde{r}_{j, t + 1}\right] - r_{f, t + 1} &= -\delta\left(1 + r_{f, t + 1}\right)\operatorname{cov}_t\left[\frac{a - b\widetilde{c}_{t + 1}}{a - bc_t},\widetilde{r}_{j, t + 1}\right] \\ &= \left(1 + r_{f, t + 1}\right)\frac{b\delta}{a - bc_t}\operatorname{cov}_t\left[\widetilde{c}_{t + 1}, \widetilde{r}_{j, t + 1}\right]\end{align}$$
My biggest question is where does the "$a$" go in the numerator??
Remember that if $X,Y$ are any random variables then $$ \textrm{cov}(X,Y)=\mathbb E(XY)-(\mathbb EX)(\mathbb EY), $$ by definition. From this definition we see a few things:
If $X=c$ is a constant (non-random) then $\textrm{cov}(c,Y)=\mathbb E(cY)-c\mathbb EY=0$.
If $a$ is any constant (non-random) then $\textrm{cov}(aX,Y)=a\textrm{cov}(X,Y)$.
In general $\textrm{cov}(X+Z,Y)=\textrm{cov}(X,Y)+\textrm{cov}(Z,Y)$.
In your question, we are going to use all three of these properties to simplify the expression:
a. Write $$ \frac{a-b\widetilde c_{t+1}}{a-bc_t}=\frac{a}{a-bc_t}-\frac{b\widetilde c_{t+1}}{a-bc_t}, $$ and apply (3) to the covariance of the sum.
b. Presumably $a,b,$ and $c_t$ are non-random in the context of your calculation, so we can apply (1) to get rid of the first term so only the second remains.
c. Use (2) to pull out the constant factors from the remaining term.