To solve this problem, we start by understanding that two vectors are perpendicular (denoted by $⊥$) if their dot product equals zero. Given vectors $\overrightarrow{a}=({1,2}), \overrightarrow{b}=({-2,1}),$ and $\overrightarrow{c}=({2,t})$, we are told that $({\overrightarrow{a}+2\overrightarrow{b}})⊥\overrightarrow{c}$. This means we need to find the value of $t$ that satisfies this condition.

First, let's calculate $\overrightarrow{a}+2\overrightarrow{b}$:
\begin{align*}
\overrightarrow{a}+2\overrightarrow{b} &= ({1,2})+2\cdot({-2,1}) \\
&= ({1,2})+({-4,2}) \\
&= ({1-4,2+2}) \\
&= ({-3,4}).
\end{align*}

Given that $({\overrightarrow{a}+2\overrightarrow{b}})⊥\overrightarrow{c}$, we apply the condition for perpendicularity, which is the dot product equals zero:
\begin{align*}
\overrightarrow{a}+2\overrightarrow{b} \cdot \overrightarrow{c} &= 0 \\
({-3,4}) \cdot ({2,t}) &= 0 \\
-3\cdot 2 + 4\cdot t &= 0 \\
-6 + 4t &= 0.
\end{align*}

Solving for $t$:
\begin{align*}
4t &= 6 \\
t &= \frac{6}{4} \\
t &= \frac{3}{2}.
\end{align*}

Therefore, the value of $t$ that satisfies the given condition is $\boxed{\frac{3}{2}}$.