I would like to contribute with a similar answer, which follows Bargmann's original article on Projective Representations.
Bargmann's approach:
In this context, we mean by $x,v,a$ vectors and by $R$ a rotation matrix. Consider the Schrödinger equation in three dimensions:
\begin{equation}
i \frac{\partial \psi(x,t)}{\partial t}+\frac{1}{2m} \Delta \psi(x,t)= V \psi(x,t).
\end{equation}
Let's look at the same state from another frame of reference, which is given by the Galilean transformations as follows:
\begin{equation}\label{Galilei}\tag{1}x'^{j}= \sum_{k=1}^3 \limits R^{jk}x^k+v^jt+a^j;\end{equation}
\begin{equation}\label{Galilei1}\tag{2} t'=t+b. \end{equation}
In the above transformations, $R$ are rotations, there are $3$ of them along each axis, $v$ represents Galilean boosts, which are also $3$, $a$ represents translations in space. Since we assume the space to be $3$-dimensional, there are 3 translations $a$. Finally, there is one translation in time, parametrized by $b$. Therefore, we have found that the Galilean group is a $10$-dimensional Lie group. In the following, we want our quantum theory to be invariant under this symmetry group, therefore we have to represent it on the Hilbert space. Two physical states are equivalent, if the corresponding wavefunctions are equal up to a phase. This means, that we look at projective representations of the Galilean group on the Hilbert space. This is also extremely important, because it tells us that if we want Galilean covariance to be satisfied we must have complex wave functions. It is not a choice of description as in electrodynamics. Writing this out explicitly, we have:
\begin{equation}\tag{2}\label{phase}
\psi(x,t)=e^{if(x',t')} \psi'(x',t').
\end{equation}
where $(x',t')$ depend on $(x,t)$ according to \ref{Galilei}, \ref{Galilei1}. We know that $\psi$ satisfies the Schrödinger equation. We want to determine $f$ such that $\psi'(x',t')$ satisfies it too:
\begin{equation}\tag{3}\label{schrodingertransformed}
i \frac{\partial \psi(x,t)}{\partial t}+ \frac{1}{2m} \Delta \psi(x,t)= V \psi(x,t).
\end{equation}
Since $f(x',t')$ is yet unknown, this restriction, namely the covariance has to determine it. Let's see how we could rewrite the above equation in terms of $x',t'$. To do that, first consider the coordinate transformations. They imply the following change of partial derivatives:
\begin{equation}\tag{4}\label{galileiderivative}
\begin{aligned}
\nabla_i= \frac{\partial}{\partial x^i}&=\sum_{j=1}^{3} \limits\underbrace{\frac{\partial x^{'j}}{\partial x^i}}_{=R^{ji}} \frac{ \partial}{\partial x^{'j}}+ \underbrace{\frac{ \partial t'}{\partial x^i}}_{=0} \frac{\partial}{\partial t'}=\sum_{j=1}^{3} \limits R_{ji} \frac{\partial}{\partial x^{'j}}=\sum_{j=1}^{3} \limits (R_{ij})^{T} \frac{\partial}{\partial x^{'j}}:=R^{-1} \nabla_i ' ;\\
\frac{\partial}{\partial t}&=\sum_{i=1}^{3}\underbrace{\frac{\partial x^{'i}}{\partial t}}_{=v^i} \frac{\partial}{\partial x^{'i}}+ \underbrace{\frac{\partial t'}{\partial t}}_{=1} \frac{\partial}{\partial t'}=\sum_{i=1}^{3} \limits v^i \frac{\partial}{\partial x{'i}}+ \frac{\partial}{\partial t'}:= v \cdot \nabla ' + \frac{\partial}{\partial t'}.
\end{aligned}
\end{equation}
We also know that the Laplacian is rotationally invariant. Since the only part in the coordinate transformation for $x'$, which contains $x$ is the rotational part, we can conclude that the Laplacian is invariant under the Galilean transformations aswell.
Therefore, we arrive at the following equation:
\begin{equation}
\left( i \frac{\partial}{\partial t'} +i v \cdot \nabla ' +\frac{1}{2m} \Delta ' - V\right) \left( e^{i f(x',t')} \psi'(x',t') \right).
\end{equation}
Before we do the product rule, let us recall a vector calculus identity, which will produce us an extra 'unintuitive' term:
\begin{equation}
\Delta(f_1 f_2)=(\Delta f_1) f_2+ f_1 (\Delta f_2)+2 (\nabla f_1) \cdot (\nabla f_2).
\end{equation}
To expand this, we have to use the product rule:
\begin{equation*}
i \underbrace{\frac{\partial}{\partial t'} \left(e^{if(x',t')} \right)}_{=i \frac{\partial{f(x',t')}}{\partial t'} e^{if(x',t')}} \psi'(x',t') +i e^{if(x',t')} \frac{\partial}{\partial t'} \psi'(x',t') + i\underbrace{(v \cdot \nabla') \left(e^{if(x',t')} \right)}_{=(iv \cdot \nabla ' f(x',t')) e^{if(x',t')}} \psi'(x',t')+ie^{if(x',t')} (v \cdot \nabla ')(\psi'(x',t'))
\end{equation*}
\begin{equation*}
+\frac{1}{2m}\left( \underbrace{\Delta' e^{if(x',t')}}_{\text{term 1}} \right) \psi'(x',t') +\frac{1}{2m}e^{if(x',t')} ( \Delta' \psi'(x',t'))+\frac{2}{2m}\left(\underbrace{\nabla' e^{if(x',t')}}_{=(i \nabla' f(x',t'))e^{if(x',t')}} \right) \cdot (\nabla ' \psi'(x',t')) -V e^{if(x',t')} \psi'(x',t')=0.
\end{equation*}
With term 1 we still have plenty of work to do. Let's compute it explicitly:
\begin{equation*}
\Delta ' e^{if(x',t')}= \nabla' \cdot \left( \nabla ' e^{if(x',t')} \right)= \nabla' \cdot \left( e^{if(x',t')} (i \nabla ' f(x',t')) \right)
\end{equation*}
\begin{equation*}
= e^{if(x',t')} \nabla'\cdot (i \nabla' f(x',t'))+ i \underbrace{\left(\nabla ' e^{if(x',t')} \right)}_{= (i \nabla' f(x',t'))e^{i f(x',t')}} \cdot \nabla' f(x',t')
\end{equation*}
\begin{equation*}
=i \left(\nabla'^2 f(x',t') \right) e^{if(x',t')} +i \nabla ' f(x',t') (i \nabla' f(x',t'))e^{if(x',t')}
\end{equation*}
\begin{equation*}
=\left(i \nabla^{'2} f(x',t')-\left(\nabla ' f(x',t') \right)^2 \right) e^{if(x',t')}.
\end{equation*}
So, we can simplify our expression after the product rule as follows:
\begin{equation*}
\begin{aligned}
& \underbrace{-\frac{\partial f(x',t')}{\partial t'} e^{if(x',t')} \psi'(x',t')}_{1} + \underbrace{ie^{if(x',t')} \frac{\partial \psi'(x',t')}{\partial t'}}_{3}-\underbrace{v \cdot (\nabla' f(x',t'))e^{if(x',t')} \psi'(x',t')}_{1}+\underbrace{e^{if(x',t')}(iv \cdot \nabla ') (\psi'(x',t')}_{2}\\
& +\underbrace{\frac{1}{2m} \left(i \nabla'^2 f(x',t') - \left(\nabla' f(x',t')\right)^2 \right)e^{if(x',t')} \psi'(x',t')}_{1}+ \underbrace{\frac{1}{2m} e^{if(x',t')}\left(( \Delta' \psi'(x',t') \right) -V e^{-if(x',t')} \psi'(x',t')}_{3} \\
&+ \underbrace{\frac{1}{m}\left( i \nabla' f(x',t') \right) e^{if(x',t')} \cdot (\nabla ' \psi'(x',t'))}_{2}=0.
\end{aligned}
\end{equation*}
Regrouping the terms in the groups $1,2,3$ yields:
\begin{equation*}
\left(\underbrace{-\frac{\partial f(x',t')}{\partial t'}+\frac{i}{2m} \nabla^{'2}f(x',t')- \frac{1}{2m}\left(\nabla' f(x',t') \right)^2 -v \cdot \nabla' f(x',t')}_{\text{condition 1}}\right) e^{i f(x',t')} \psi'(x',t')
\end{equation*}
\begin{equation*}
+ i\left(\underbrace{v +\frac{1}{m} \nabla' f(x',t')}_{\text{condition 2}} \right)e^{if(x',t')} \cdot \left( \nabla ' \psi'(x',t') \right) + e^{if(x',t')}\underbrace{\left( i\frac{ \partial}{\partial t'}+\frac{1}{2m} \Delta '-V \right) \psi'(x',t')}_{\text{Schrödinger}}=0.
\end{equation*}
The last part is the Schrödinger equation for $\psi'(x',t')$. So we want to kill all the other terms in order for this to be equal to 0 on the RHS to guarantee covariance. We get the following conditions on $f$ to assure covariance:
\begin{equation}\label{conditions1} \tag{4}
v+\frac{1}{m} \nabla ' f(x',t')=0 ;
\end{equation}
\begin{equation}\label{conditions2} \tag{5}
- \frac{\partial f(x',t')}{\partial t'}+\frac{i}{2m}\nabla^{'2}f(x',t')- \frac{1}{2m}\left(\nabla' f(x',t') \right)^2 -v \nabla' f(x',t')=0.
\end{equation}
The above system can be easily integrated to obtain:
\begin{equation}
f(x',t')=-mv \cdot x'+\frac{1}{2}mv^2 t' + C.
\end{equation}
To check that this is indeed the case, let's substitute back into \ref{conditions1},\ref{conditions2}. To do this, compute all the objects appearing in the conditions:
\begin{equation}
\nabla' f(x',t')=-mv;
\end{equation}
\begin{equation}
\frac{\partial f(x',t')}{\partial t'}=\frac{1}{2}mv^2;
\end{equation}
\begin{equation}
\nabla^{'2}f(x',t')=0.
\end{equation}
Therefore, we have:
\begin{equation*}
v+\frac{1}{m}(-mv)=v+(-v)=v-v=0.
\end{equation*}
\begin{equation*}
-\frac{1}{2} mv^2+\underbrace{\frac{-i}{2m}0}_{=0}-\underbrace{\frac{1}{2m} m^2 v^2}_{=\frac{1}{2} mv^2} - \underbrace{v(-mv)}_{=mv^2}=0 \implies -\frac{1}{2} mv^2 -\frac{1}{2} mv^2 + mv^2=0
\implies 0=0.
\end{equation*}
This shows that the above function $f(x',t')$ solves the system. So the wavefunctions transform under a Galilean transformation as:
\begin{equation}
\psi(x,t)=e^{i f(x',t')}\psi'(x',t') \implies \psi'(x',t')=e^{-if(x',t')} \psi(x,t).
\end{equation}
Writing out $f(x',t')$ explicitly:
\begin{equation}\label{representation}
\psi'(x',t')=e^{i \left( mv \cdot x' - \frac{1}{2}mv^2t'+C \right)} \psi(x,t).
\end{equation}
Or equivalently:
\begin{equation}\label{representation1}
\psi'(Rx+vt+a,t+b)= e^{i \left(mv \cdot (Rx+vt+a)- \frac{1}{2} mv^2(t+b) +C \right)} \psi(x,t).
\end{equation}
Since we have derived the most general case, let's discuss the textbook examples. One widely known textbook example, for example the one discussed by Merzbacher in his Quantum Mechanics book is the Galilean boost invariance. To do that, we have to set $R=1,a=b=0$:
\begin{equation}
\psi'(x+vt,t)=e^{i \left( mv \cdot(x+vt) - \frac{1}{2} mv^2t \right)} \psi(x,t).
\end{equation}
Or explicitly as it appears in the book:
\begin{equation}
\psi'(x,t)=e^{im \left( v \cdot x - \frac{1}{2}v^2t \right)} \psi(x-vt,t).
\end{equation}
References:
Valentine Bargmann, On Unitary Ray Representations of Continuous Groups, Annals of Mathematics, 1954.
Eugen Merzbacher, Quantum Mechanics, Third Edition
Special thanks goes to Valter Moretti, who helped me spot a mistake in my computation back then when I was doing it and struggled to get the equivalence of the results.
