Gauss' contiguous relations provide a basis for finding a linear relationship between three functions of the form ${}_2F_1(a+k, b+l, c+m, z)$, henceforth $\mathbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right)$. Various papers have been published on calculating these relationships; I'm using the notation of Vidūnas's paper Contiguous relations of hypergeometric series:
$$\textbf{F}\left(\begin{matrix}a+k, b+l \\ c+m\end{matrix}\right) = \textbf{P}(k,l,m) \textbf{F}\left(\begin{matrix}a, b \\ c\end{matrix}\right) + \textbf{Q}(k,l,m) \textbf{F}\left(\begin{matrix}a+1, b \\ c\end{matrix}\right)$$
Vidūnas recommends finding the coefficients by first chaining up $k$ with his formula (9), but a naïve approach runs into bootstrap problems (as I discovered by trying it). Once $\textrm{Q}(2,0,0)$ is found, I think the approach works, but here we have such small shifts that we can do directly.
Rephrasing his (2), (3), (4) in terms of $\textbf{Q}$ we obtain
$$\mathbf{Q}(k+1,l,m) = \tfrac{(2a+2k-c-m+(b+l-a-k)z)}{(a+k)(1-z)}\mathbf{Q}(k,l,m) + \tfrac{(c+m-a-k)}{(a+k)(1-z)}\mathbf{Q}(k-1,l,m)$$
$$\mathbf{Q}(k,l,m-1) = \tfrac{a+k}{c+m-1} \mathbf{Q}(k+1,l,m) + \tfrac{c+m-a-k-1}{c+m-1} \mathbf{Q}(k,l,m)$$
$$\mathbf{Q}(k,l+1,m) = \tfrac{a+k}{b+l} \mathbf{Q}(k+1,l,m) + \tfrac{b+l-a-k}{b+l}\mathbf{Q}(k,l,m)$$
and by repeated application of these I calculate that
$$\mathbf{Q}(0,2,-1) = a \tfrac{(a+b-c+1)(a+b-c+2) + (c-a-1)(2a+b-c+1)(1-z) + (b-a+1)(c-a-1)(1-z)^2}{(c-1)b(b+1)(1-z)^2} \\
$$
$$\mathbf{P}(0,2,-1) = (c-a-1) \tfrac{a(a+b-c+2) + a(-2a+b+c-1)(1-z) + (b-a)(b-a+1)(1-z)^2}{(c-1)b(b+1)(1-z)^2}$$
Substituting your parameters $a=\tfrac12$, $b=2t$, $c=1-t$, $z=4$ I get respectively $0$ and $1$, which establishes the periodicity of your $f$.
To search systematically for scale-periodic hypergeometric functions, I propose the following approach:
- Pick $k$, $l$, $m \in \mathbb{Z}$ (taking into account symmetries for efficiency).
- Compute $\mathbf{Q}(k,l,m)$ and $\mathbf{P}(k,l,m)$.
- Substitute $a = a_0 + kt$, $b = b_0 + lt$, $c = c_0 + mt$ into $\mathbf{P}(k,l,m) = A$ and $\mathbf{Q}(k,l,m) = 0$. Equate coefficients of $t$ in each to get a collection of polynomial constraints over $a_0,b_0,c_0,z,A$. Use Gröbner bases or other techniques to solve these constraints.
To take the obvious example of $k=0$, $l=2$, $m=-1$, and (for simplicity) applying the substitution $x = 1-z$, $\textbf{Q}(0,2,-1) = 0$ gives
$$
a_0 (9-3x-2x^2) = 0 \\
6a_0+6b_0-6c_0+9+(-5a_0-b_0+4c_0-4)x+(-a_0-b_0+2c_0-3)x^2 = 0 \\
(a_0+b_0-c_0+1)(a_0+b_0-c_0+2) +
(-a_0+c_0-1)(2a_0+b_0-c_0+1)x +
(-a_0+b_0+1)(-a_0+c_0-1)x^2 = 0
$$
Ignoring the trivial $a_0 = 0$, we see that $x \in \{-3, \tfrac32\}$.
Case $x=-3$
$a_0 = \tfrac12$ drops out of the next constraint, leaving the final constraint to reduce to $(b_0+2c_0-3)(b_0+2c_0-2) = 0$. If $b_0+2c_0-2 = 0$ then we're going to recover your $f$, so consider instead $b_0+2c_0-3 = 0$. This turns out to have no solutions: substituting into $\mathbf{P}$ and expanding we get a contradiction in the coefficients of $t^0$.
Case $x = \tfrac32$
Comparing coefficients for $t^3$ in $\mathbf{P}$ gives $A = 1$; subsequently comparing coefficients for $t^2$ in $\mathbf{P}$ gives $a_0 = 0$, so there is no non-trivial solution.
I therefore withdraw my blithe assertion in an earlier comment that question (2) is easily answered in the affirmative; it remains plausible, but with larger shifts we're going to have more coefficients of $t$ to impose constraints on a fixed number of variables and so there might not be very many sets of shifts which admit solutions.
