Andrey Pressley states in his book Elementary Differential Geometry (pp. 29) that
Suppose that $\gamma$ is a unit-speed curve in $\mathbb{R}^2$. As the parameter $t$ of $\gamma$ changes to $t + \delta t$, the curve moves away from its tangent line at $\gamma(t)$ by a distance $(\gamma(t + \delta t) - \gamma(t))\cdot \mathbb{n}$, where $\mathbb{n}$ is the unit vector perpendicular to the tangent vector $\dot{\gamma}(t)$ of $\gamma$ at the point $\gamma(t)$
The author does not provide any intermediary steps for this conclusion. My question is is that why does the following interpretation lead to the same result: As we are interested in the deviation from the tangent $\dot{\gamma}(t)$, we can simply i.) project the normal vector onto the different vector $\gamma(t + \delta t) - \gamma(t)$, ii.) find its norm. Combining these two steps gives $d = \left|\left|\mathrm{proj}_{\gamma(t + \delta t) - \gamma(t)}(\mathbb{n})\right|\right| = \left|\left|(\gamma(t + \delta t) - \gamma(t))(\frac{<\gamma(t + \delta t) - \gamma(t), n>}{\left|\left|\gamma(t + \delta t) - \gamma(t)\right|\right|\left|\left|\mathbb{n}\right|\right|})\right|\right| = \left|\left|<\gamma(t + \delta t) - \gamma(t), n>)\right|\right| = (\gamma(t + \delta t) - \gamma(t))\cdot \mathbb{n}$
I am confused by this interpretation as to me it would make more sense to project the difference vector $\gamma(t + \delta t) - \gamma(t)$ onto the normal vector, rather than the other way around, as we'd like to know how much we have moved away from the tangent.
Your formula for the orthogonal projection is wrong. The orthogonal projection of a vector $v$ onto a vector $w$ is $P_wv = (v, \frac{w}{\lVert w \rVert})\frac{w}{\lVert w \rVert} = \frac{1}{\lVert w \rVert^2}(v, w)$.
It's not clear what the author is saying. Let $x$ be the closest vector in $\gamma(t) + \text{span}(\gamma'(t))$ to $\gamma(t + \delta t)$. He might mean that $$\lVert \gamma(t + \delta t) - x \rVert = |(\gamma(t + \delta t) - \gamma(t)) \cdot n|.$$ We have $$\gamma(t + \delta t) - x= \gamma(t + \delta t) - \gamma(t) - v,$$ where $v$ is the closest vector in $\text{span}(\gamma'(t))$ to $\gamma(t + \delta t) - \gamma(t)$. From linear algebra, we know that $v$ is the orthogonal projection of $\gamma(t + \delta t) - \gamma(t)$ onto $\text{span}(\gamma'(t))$. Therefore $\gamma(t + \delta t) - \gamma(t) - v$ is the orthogonal projection of $\gamma(t + \delta t) - \gamma(t)$ onto $\text{span}(\gamma'(t))^{\perp} = \text{span}(n)$, so $\gamma(t + \delta t) - \gamma(t) - v = (\gamma(t + \delta t) - \gamma(t)) \cdot n$.