Let $\mathbb{F}[X]$ be the polynomial ring with one variable.
$V$ can be regarded as an $\mathbb{F}[X]$-module by defining $Xv = T(v)$ for every $v \in V$. $\mathbb{F}[X]$-submodules of $V$ are none other than $T$-invariant subspaces of $V$.
Let $K = \mathbb{F}[X]/(p_t)$.
Since $p_t$ is irreducible, $K$ is a field.
Since $p_t V = 0$, $V$ can be regarded as a $K$-module.
Let $W$ be a $T$-invariant subspace of $V$.
$W$ can be regarded as a $\mathbb{F}[X]$-submodule of $V$.
Since $p_t W = 0$, $W$ can be regarded as a $K$-submodule of $V$.
Hence there exists a $K$-submodule $W'$ such that $V = W \oplus W'$.
Since $W'$ is a $\mathbb{F}[X]$-submodule, it is $T$-invariant.
This completes the proof.
$\mathbb Z_{180}\simeq \mathbb Z_4\oplus\mathbb Z_9\oplus\mathbb Z_5$ (here I've used repeatedly that $\mathbb Z_{mn}\simeq\mathbb Z_m\oplus\mathbb Z_n$ whenever $(m,n)=1$). This shows that for this group the $2$-primary component is $\mathbb Z_4$, the $3$-primary component is $\mathbb Z_9$, and the $5$-primary component is $\mathbb Z_5$.
For the invariant factor decomposition: $180$ is the only invariant factor of this group, since the elementary divisors are $2^2$, $3^2$, and $5$. It can't be true that $\mathbb Z_{180}\simeq\mathbb Z_6\oplus\mathbb Z_{30}$: if this holds, then $30$ will kill every element of $\mathbb Z_{180}$, which is obviously false. The same can be said for the other decompositions.
Edit. The OP asked in the comments about similar decompositions for the group $\mathbb Z_{24}\oplus\mathbb Z_{30}\oplus\mathbb Z_{20}.$ In this case the elementary divisors are $2^3$, $3$, $2$, $3$, $5$, $2^2$, $5$. This shows that the $2$-primary component of this group is $\mathbb Z_8\oplus\mathbb Z_2\oplus\mathbb Z_4$, the $3$-primary component is $\mathbb Z_3\oplus\mathbb Z_3$, and the $5$-primary component is $\mathbb Z_5\oplus\mathbb Z_5$.
The invariant factor decomposition can be found by using the elementary divisors: the greatest invariant factor is $120=2^3\cdot 3\cdot 5$, the next one is $60=2^2\cdot 3\cdot 5$, and the last one is $2$. Thus we have the following invariant factor decomposition: $\mathbb Z_{2}\oplus\mathbb Z_{60}\oplus\mathbb Z_{120}.$
Best Answer
It is not necessarily the case that all invariant subspaces can be written as a sum of $W_i$. For instance, consider $$ T = \left[\begin{array}{cc|cc} 0&1&0&0\\0&0&0&0\\ \hline 0&0&0&1\\0&0&0&0 \end{array} \right] $$ with cyclic decomposition $$W_1 = \{(x_1,x_2,0,0):x_1,x_2 \in \Bbb F\}, \quad W_2 = \{(0,0,x_3,x_4):x_3,x_4 \in \Bbb F\}. $$ The invariant subspace $\ker(T) = \{(x_1,0,x_3,0): x_1,x_3 \in \Bbb F\}$ cannot be written as a direct sum of $W_i$.