Is there a reason we need to make a logistic regression linear using the logit? My understanding is that we use the logit function to convert the sigmoidal curve of a logistic regression to be linear.
As a result, we go from a curve modeled as P = ea+bX / (1 + ea+bX) to one that can be modeled linearly as logit for p = a + bX.
Why is this important? (why do we need to do this?)
One suggestion I found was that this conversion to log odds shifts the model from being probability based to being a likelihood-based model. IS this true? If so, what value does that hold?
Summary: why do we transform a logistic regression equation to be linear?

It's a bit hard to know what you're asking and how to answer it without completely explaining logistic regression, but the sigmoid curve
$$
P(Y=1|X) = \frac{e^{a+bx}}{1+e^{a+bx}}
$$
and the linear logit regression equation
$$
\text{logit}(P(Y=1|X))=a+bx
$$
are equivalent ways of writing the same relationship. There is no transformation. The former makes it clear that we are modeling a probability using a sigmoid curve; the latter makes it clear that we are modeling the logit of the probability using a linear function. The two equations emphasize different parts of the relationship, but they express identical information.
To see how, note the definition of the logit:
$$
\text{logit}(p) = \text{log}\left(\frac{p}{1-p}\right) = \text{log}(p)-\text{log}(1-p)
$$
Inserting the sigmoid model for $p$, we get
\begin{align}
\text{logit}(P(Y=1|X)) &= \text{log}\left(\frac{e^{a+bx}}{1+e^{a+bx}}\right)-\text{log}\left(1-\frac{e^{a+bx}}{1+e^{a+bx}}\right) \\
&= \text{log}\left(\frac{e^{a+bx}}{1+e^{a+bx}}\right)-\text{log}\left(\frac{1}{1+e^{a+bx}}\right) \\
&= \text{log}\left(e^{a+bx}\right) - \text{log}\left(1+e^{a+bx}\right) + \text{log}\left(1+e^{a+bx}\right) \\
&= \text{log}\left(e^{a+bx}\right) \\
&= a + bx
\end{align}
