I would like to see an example of a fiber bundle $\mathbb{T}^2 \to X \xrightarrow{\pi} B$ whose fibers are tori $\mathbb{T}^2:=(S^1)^2$ and whose vertical tangent bundle $\ker(d\pi) \to X$ is not decomposable? Which homotopy groups does one need to look at to construct such an example?
(This problem has been edited using inputs from the comments.)
Put $T=\{(z_0,z_1,z_2)\in(S^1)^3:z_0z_1z_2=1\}$, so $T$ is homeomorphic to $S^1\times S^1$ and has an obvious action of the symmetric group $\Sigma_3$. The action of $\Sigma_3$ on $H_1(T;\mathbb{R})$ is indecomposable, and this homology group can also be identified with the tangent space to $T$ at the identity element.
Now put $F=\{(u_0,u_1,u_2)\in\mathbb{C}^3:u_i\neq u_j\text{ for } i\neq j\}$. This also has a natural action of $\Sigma_3$, which is free. Put $E=(T\times F)/\Sigma_3$ and $B=F/\Sigma_3$, so we have a fibre bundle $T\to E\to B$. This will have the required indecomposability. Note that the identity element of $T$ gives a section of the projection $E\to B$.
The cohomology of $F$ is a well-known example. For $i\neq j$ we can define $\pi_{ij}\colon F\to\mathbb{C}\setminus\{0\}$ by $\pi_{ij}(z)=z_j-z_i$, and by pulling back the generator of $H^1(\mathbb{C}\setminus\{0\})$ we get an element $a_{ij}\in H^1(F)$. These satisfy $a_{ij}=a_{ji}$ and $a_{ij}^2=0$ and the Arnol'd relation $a_{ij}a_{jk}+a_{jk}a_{ki}+a_{ki}a_{ij}=0$. It follows that $\{1,a_{01},a_{02},a_{12},a_{01}a_{02},a_{01}a_{12}\}$ is a basis for $H^*(F)$. From this description you can calculate $H^*(B)$ and $H^*(E)$ using the Serre spectral sequence. With rational coefficients this just degenerates to $H^*(B;\mathbb{Q})=H^*(F;\mathbb{Q})^{\Sigma_3}=\mathbb{Q}$ and $H^*(E;\mathbb{Q})=H^*(T\times F;\mathbb{Q})^{\Sigma_3}$. It can't be too hard to calculate that last ring explicitly, but I have not done so.
If you prefer to work with homotopy groups, it is standard that $\pi_1(B)$ is the braid group $Br_3$. This has a standard surjective homomorphism to $\Sigma_3$, whose kernel is the pure braid group $PBr_3$, which can be identified with $\pi_1(F)$. It is also known that $\pi_i(F)=\pi_i(B)=\pi_i(T)=0$ for $i>1$. The map $PBr_3\to\Sigma_3$ gives an action of $PBr_3$ on $\pi_1(T)\simeq\mathbb{Z}^2$, using which we can form the semidirect product $PBr_3\ltimes\mathbb{Z}^2$. Using the split fibre bundle $T\to E\to B$ we can identify $\pi_1(E)$ with this semidirect product, and also see that $\pi_i(E)=0$ for $i>1$.
