If we look at he Klein bottle $K$ as a $S^1$-fiber bundle over $S^1$, we can apply the long exact sequence in Homotopy for fibers. $$\pi_2(S^1)\rightarrow\pi_2(K)\rightarrow\pi_2(S^1)\rightarrow\pi_1(S^1)\rightarrow\pi_1(K)\rightarrow\pi_1(S^1)\rightarrow\dots$$ We can then easily see directly that $\pi_2(K)$ vanishes.
In particular we get the short exact sequence $$0\rightarrow\pi_1(S^1)\rightarrow\pi_1(K)\rightarrow\pi_1(S^1)\rightarrow 0.$$Is that enough to deduce something about $\pi_1(K)$?
I know it is not the optimal approach but I am interestet in how much information i get out of the fiber bundle.
You can deduce for example that $\pi_1(K)$ is generated by $x$ the image of the generator of $\pi_1(S^1)$ by $f:\pi_1(S^1)\rightarrow\pi_1(K)$, and $y$ the element of $\pi_1(K)$ whose image by $g:\pi_1(K)\rightarrow \pi_1(S^1)$ is the generator of $\pi_1(S^1)$.
To see this, let $z\in\pi_1(K)$, there exists $n\in\mathbb{Z}$ such that $g(z)=g(y^n)$, we deduce that $g(zy^{-n})=1$ and $zy^{-n}=x^m$, we conclude that $z=x^my^n$.