Noether's Theorem and Lagrangian symmetries

If I make the transformation

$$x_i \rightarrow x_i' = x_i + \delta x_i,$$

I find that the Lagrangian $L = L(x_i,\dot{x_i})$ transforms as

$$L \rightarrow L' = L + \frac{\partial L}{\partial x_i} \delta x_i + \frac{\partial L}{\partial \dot{x_i}} \delta \dot{x}_i \\ = L + \frac{\partial L}{\partial x_i} \delta x_i + \frac{\partial L}{\partial \dot{x_i}} \frac{d}{dt}\delta x_i \\ =L + \frac{\partial L}{\partial x_i} \delta x_i + \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x_i}} \delta x_i\bigg) - \bigg(\frac{d}{dt}\frac{\partial L}{\partial \dot{x_i}}\bigg) \delta x_i \\ = L + \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x_i}} \delta x_i \bigg).$$

For a symmetry, we demand $L'=L$, which means

$$\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x_i}} \delta x_i \bigg) = 0. \quad (\mathrm{1})$$

This is just Noether's theorem. However, I have also read in Goldstein, and other Classical mechanics texts, that if we transform the Lagrangian in such a way that it changes as

$$L \rightarrow L' = L + \frac{d}{dt} F(x) \quad (\mathrm{2}),$$

$L'$ still extremises the action for the same path. So, surely for symmetry I do not have to demand the condition as given in $(\mathrm{1})$ because it is a total time derivative like $(\mathrm{2})$. Is Noether's theorem overly restrictive about Lagrangian transformations? Can I derive conserved quantities even if I allow the Lagrangian to change by a total time derivative?

I think the problem with your statement is the definition of symmetry.If you consider the family of trasformations $$x'_i \rightarrow x_i + \epsilon K_i(\vec{x},\dot{\vec{x}})$$ Then you can demand for symmetry that: $$\dfrac{dL'}{d\epsilon}\Bigr|_{\epsilon=0}=0$$ So that a symmetry of the Langrangian is a transformation that leaves it unchanged at first order with respect to $\epsilon$.You can derive the conserved quantity as follows from Taylors theorem: $$L'(x_i',\dot{x_i'})=L'(x_i+\epsilon K,\dot{x_i}+\epsilon \dot{K})=L(x_i,\dot{x_i}) +\dfrac{\partial L}{\partial x_i}\epsilon K_i +\dfrac{\partial L}{\partial \dot{x_i}}\epsilon \dot{K_i}+\mathcal{O}(\epsilon^2)$$ From there by applying the condition of the symmetry and using the Euler-Lagrange equations you get your desired result: $$\dfrac{\partial L}{\partial x_i}K_i +\dfrac{\partial L}{\partial \dot{x_i}} \dot{K_i} = \dfrac{d}{dt}\left( \dfrac{\partial L}{\partial \dot{x_i}}\right) K_i + \dfrac{\partial L}{\partial \dot{x_i}} \dot{K_i} = \dfrac{d}{dt}\left( \dfrac{\partial L}{\partial \dot{x_i}} K_i\right)=0$$
Combining your calculation of $L'-L$ with th claim in Eq. (2), the conserved quantity is $\dfrac{\partial L}{\partial \dot{x}_i}\delta x_i-F\left( x\right)$. Similarly, $F$ must in general be subtracted from the conserved charge obtained in SpecialM's answer.