We can compute the generating function directly from the definition
$$U_n(\cos \theta) = \frac{\sin (n+1)\theta}{\sin \theta}.$$
Write $z = e^{i \theta}$. Then the RHS is $\frac{z^{n+1} - z^{-n-1}}{z - z^{-1}}$, and so the exponential generating function is
$$\begin{align*} \sum_{n \ge 0} U_n(\cos \theta) \frac{t^n}{n!} &= \sum_{n \ge 0} \frac{z^{n+1} - z^{-n-1}}{z - z^{-1}} \frac{t^n}{n!} \\
&= \frac{ze^{tz} - z^{-1} e^{tz^{-1}}}{z - z^{-1}} \\
&= \frac{(\cos \theta + i \sin \theta) e^{t(\cos \theta + i \sin \theta)} - (\cos \theta - i \sin \theta) e^{t(\cos \theta - i \sin \theta)}}{2i \sin \theta} \\
&= e^{t \cos \theta} \frac{(\cos \theta + i \sin \theta) e^{it \sin \theta} - (\cos \theta - i \sin \theta) e^{-it \sin \theta}}{2i \sin \theta} \\
&= e^{t \cos \theta} \frac{2i \cos \theta \sin (t \sin \theta) + 2i \sin \theta \cos (t \sin \theta)}{2i \sin \theta} \\
&= e^{t \cos \theta} \left( \cos (t \sin \theta) + \cot \theta \sin (t \sin \theta) \right) \\
&= e^{tx} \left( \cos (t \sqrt{1 - x^2}) + \frac{x}{\sqrt{1 - x^2}} \sin (t \sqrt{1 - x^2}) \right) \end{align*}$$
where $x = \cos \theta$. To convert to the hyperbolic sine and cosine form given on Wikipedia write $\sqrt{1 - x^2} = i \sqrt{x^2 - 1}$ and use that $\cos (it) = \cosh t$ and $\sin it = i \sinh t$; I don't know why Wikipedia writes it that way.
For some corroboration, in Clemente Cesarano's Identities and generating functions for Chebyshev polynomials, you can find an identity for $\sum U_{n-1}(x) \frac{t^n}{n!}$ (Proposition 2) which appears to be equivalent to this one although one needs to differentiate it in $t$.
I was too focused on ${_2F_1}$ functions that I missed the very obvious fact that it is not the only hypergeometric function.
This was the first time ever that I had dealings with Gamma Function, double factorials and hypergeometric function. So, I'll lay out all steps, some of which must be too obvious for veteran mathematicians.
Firstly, $(1)$ is full of falling factorials, which are not great for hypergeometric functions, so I reordered the sum by substituting $q-i \rightarrow i$, i.e.,
$$
\tag{2}
S =2\times(-1)^q\sum_{i=0}^q(-1)^i \binom{q+1+i}{q-i} 2^{2i} \frac{1}{2^{N+2i+1}} \binom{N+2i+1}{\frac{N+1}{2}+i}
$$
Then, I used the binomial identity for $\binom{2z}{z}$ that I stated above and reached here.
$$
\tag{3}
S =\frac{2\times(-1)^q}{\sqrt{\pi}}\sum_{i=0}^q(-1)^i \binom{q+1+i}{q-i} 2^{2i} \frac{\Gamma(\frac{N}{2}+i+1)}{\Gamma(\frac{N+1}{2}+i+1)}
$$
Then, I expanded the binomial. It has the only falling factorial in this expression.
$$
\tag{4}
S = \frac{2\times(-1)^q}{\sqrt{\pi}}\sum_{i=0}^q(-1)^i \frac{(q+i+1)!}{(q-i)!(2i+1)!} 2^{2i} \frac{(\frac{N}{2}+i)!}{(\frac{N+1}{2}+i)!}
$$
The get rid of the falling factorial, I multiplied both the numerator and the denominator by ${q!i!}$.
$$
\tag{5}
S = \frac{2\times(-1)^q}{q!\sqrt{\pi}}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+i+1)!i!}{(2i+1)!} 2^{2i} \frac{(\frac{N}{2}+i)!}{(\frac{N}{2}+i+\frac{1}{2})!}
$$
It was kind of straightforward except for the $(2i+1)!$.
\begin{align}
\tag{6}
S &= \frac{2(q+1)(q!)(-1)^q}{q!\sqrt{\pi}}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+2)_i(1)_i}{(2i)!!(2i+1)!!} \frac{(\frac{N}{2}+i)!}{(\frac{N}{2}+i+\frac{1}{2})!}4^{i}\\
\tag{7}
&= \frac{2(q+1)(-1)^q}{\sqrt{\pi}}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+2)_i(1)_i}{2^{2i}i!(i+\frac{1}{2})!} \frac{(\frac{N}{2}+i)!}{(\frac{N}{2}+i+\frac{1}{2})!}4^{i}\\
\tag{8}
&=\frac{2(q+1)(-1)^q}{\sqrt{\pi}}\frac{\Gamma(\frac{N}{2}+1)}{\Gamma(\frac{N+3}{2})}\sum_{i=0}^q(-1)^i\binom{q}{i} \frac{(q+2)_i}{(3/2)_i} \frac{(\frac{N}{2}+1)_i}{(\frac{N+3}{2})_i}
\end{align}
I am not sure if it can be expressed as ${}_2F_1$, but I know that it can be expressed as
$$
S = \frac{2(q+1)(-1)^q}{\sqrt{\pi}}\frac{\Gamma(\frac{N}{2}+1)}{\Gamma(\frac{N+3}{2})} {}_3F_2(-q,q+2,\frac{N}{2}+1;\frac{3}{2},\frac{N+3}{2};1).
$$
During the derivation, I also understood why there is an $(n+1)$ factor leading ${}_2F_1$ for the Chebyshev Polynomial. As a buyer's remorse, I am not sure if a closed-form expression helps with anything.
Best Answer
We show the following identity is valid for $n\geq 0$: \begin{align*} \color{blue}{\sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} (-1)^k \binom{n-k}{k}~(2x)^{n-2k} =\sum_{k=0}^{n}(-2)^{k}\binom{n+k+1}{2k+1}(1 - x)^k }\tag{1} \end{align*}
The bivariate generating function of the Chebyshev polynomials of the second kind is \begin{align*} \color{blue}{u(x,y)=\frac{1}{1-2xy+y^2}=\sum_{n=0}^{\infty}U_n(x)y^n} \end{align*} In the following we consider the generating function $\pi(x,y)$ \begin{align*} \pi(x,y)=\frac{1}{1-(2+x)y+y^2}=\frac{1}{(1-y)^2-xy}\tag{2} \end{align*} which is related with $u(x,y)$ by a transformation $x\to 2x-2$: \begin{align*} \color{blue}{\pi(2x-2,y)=u(x,y)} \end{align*}
Comment:
In (3.1) we use a geometric series expansion.
In (3.2) we apply the binomial theorem.
In (3.3) we change the order of the inner sum $k\to n-k$.
In (3.4) we exchange the order of the sums.
In (3.5) we shift the index $n$ by $k$ to change $y^{n-k}\to y^n$.
In (3.6) we shift again the order of the sums.
Comment:
In (4.1) we use a geometric series expansion.
In (4.2) we use a binomial series expansion.
In (4.3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.
In (4.4) we exchange the order of the sums and we shift the index $n$ by $k$ to change $y^{n-k}\to y^n$.
In (4.5) we exchange the order of the sums.
In (4.6) we change the order of the inner sum $k\to n-k$.
In (4.7) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.
Note: This answer is a solution to chapter 2, problem 18.c in Combinatorial Identities by John Riordan.