How does this T-network op amp function as a boost filter?
You're missing the action of the op-amp.
Let's start with your transfer function so far.  We'll find out that the notation is a bit unfortunate, but bear with me.  We need to do the good ol' op-amp thing where we set \$v_- = v_+\$.  The T filter is driven by the op-amp output, and feeds the op-amp input.  So your &quot;\$v_o\$&quot; is \$v_-\$.  Similarly, your &quot;\$v_i\$&quot; is the op-amp output (hence the comment about unfortunate notation).  So let
$$H(s) = \frac{V_o}{V_i}$$
Translating to my terms, and using \$V_{out}\$ for the op-amp output, in hopes of avoiding completely hopeless difficulties in notation: \$V_- = V_{out}H(s)\$
Let \$V_+ = V_-\$.  Then \$V_+ = V_{out}H(s)\$.  Now solve for \$V_{out}\$: $$V_{out}(s) = \frac{V_+(s)}{H(s)}$$
So because the feedback is a notch, the overall circuit boosts.