Confusion on Bayes theorem Urn 1: $1$ blue ball, $3$ red balls
Urn 2: $3$ blue balls, $1$ red ball
Suppose you take a ball at random without knowing which urn you took it from. Both urns are equally likely. Using Bayes’ rule, compute the probability of a majority-blue bin after two blue balls in a row. What is this probability?
It was computed in a previous problem that the probability of drawing two blue balls in a row is $\frac{5}{16}$ The probablity of drawing from urn 2 would be $\frac{1}{2}$ and the probability of picking a blue ball from that urn is $\frac{3}{4}$. 
$\frac{\frac{1}{2}*\frac{3}{4}}{\frac{5}{16}}$
$=\frac{6}{5}$
I'm confused on where the flaw is in my computation/thinking, thank you!
 A: Presumably we select an urn, the select twice, with replacement, from that urn.  
(Because if we are selecting a different urn each time the question makes little sense, and if we are selecting without replacement from the same urn then we would certainly be selecting two blue balls from the urn with more than one ball.)
The probability of selecting two blue balls, with replacement between selections, from urn 1 is $\mathsf P(B{=}2\mid U{=}1)={(\tfrac {1}{4})}^2$
The probability of selecting two blue balls, with replacement between selections, from urn 1 is $\mathsf P(B{=}2\mid U{=}2)={(\tfrac {3}{4})}^2$
As noted, the probability of selecting each urn is equal, so what you wanted was:
$$\mathsf P(U{=}2\mid B{=}2) = \dfrac{(\tfrac 12){(\tfrac {3}{4})}^2}{(\tfrac 12){(\tfrac {3}{4})}^2+(\tfrac 12){(\tfrac {1}{4})}^2}=0.9$$
Where as you had:
$$\dfrac{(\tfrac 12){(\tfrac {3}{4})}}{\tfrac {5}{16}}$$

I'm confused on where the flaw is in my computation/thinking

You clearly have the right denominator, as $(\tfrac 12){(\tfrac {3}{4})}^2+(\tfrac 12){(\tfrac {1}{4})}^2=\tfrac 5{16}$, so it looks like you made a simple typo, and dropped the square from the numerator.   Better look on the floor.
