If $f(t)=\sum_{n=1}^{\infty} e^{-n^2t}t^a$ , then $ \lim_{t \to 0}f(t) = 0$

The limit is not always zero. It depends on $a$.

Let $\theta(t) := \sum_{n \in \mathbb{Z}}e^{-\pi n^2 t}$ for $t >0$. It is well-known (and not hard to prove via Poisson summation) that $$ \theta(t) = t^{-1/2}\theta(1/t) $$ for all $t >0$. We can write $$ f(t) = t^{a}\frac{\theta(t/\pi)-1}{2} = t^{a}\frac{\pi^{1/2}t^{-1/2}\theta(\pi/t)-1}{2}. $$ We have $\lim_{t \rightarrow 0}{\theta(\pi/t)} = 1$ and so $$ f(t) \sim t^{a}\frac{\pi^{1/2}t^{-1/2}-1}{2} \sim \frac{\pi^{1/2}}{2 }t^{a-1/2}, $$ where $g(t)\sim h(t)$ means that $g(t)/h(t) \rightarrow 1$ as $t \rightarrow 0$. In summary, for $a \in \mathbb{R}$, $$ \lim_{t \rightarrow 0}{f(t)} = \begin{cases} \infty & a < 1/2\\ \frac{\pi^{1/2}}{2 } &a =1/2\\ 0 & a>1/2 \end{cases} $$ The last case can also easily be shown via a geometric series estimate.

The question relates to the Gauss circle problem. Indeed by considering $$ \Theta(x)=\sum_{n\in\mathbb{Z}}e^{-n^2 x}$$ we have $$ \Theta(-\log z)^2 = \sum_{N\geq 0} r_2(N) z^N $$ where $ r_2(N)=\left|\left\{(a,b)\in\mathbb{Z}^2:a^2+b^2=N\right\}\right| $. By Gauss geometric estimations we have $$ \sum_{N=0}^{M}r_2(N) = \pi M + O(\sqrt{M}) $$ hence $$ \frac{\Theta^2(-\log z)}{1-z}=\frac{\pi z}{(1-z)^2}+O\left(\sum_{n\geq 0}\sqrt{n}z^n\right) $$ and $$ \Theta(-\log z)\sim\sqrt{\frac{\pi z}{1-z}} $$ as $z\to 1^-$. This implies $$ \sum_{n\geq 0}e^{-n^2 x}\sim -\frac{1}{2}+\frac{1}{2}\sqrt{\frac{\pi e^{-x}}{1-e^{-x}}}\sim -\frac{1}{2}+\frac{\sqrt{\pi}}{2\sqrt{x}} $$ as $x\to 0^+$.