Let $\lambda_1,\ldots,\lambda_k$ be the distinct eigenvalues of $\phi_1$; if the characteristic polynomial of $\phi_1$ splits (something you seem to be assuming implicitly), then we have $V = V_{\lambda_1}\oplus V_{\lambda_2}\oplus\cdots\oplus V_{\lambda_k}$ where $V_{\lambda_j} = \{x\in V\mid \text{there exists }p\text{ such that }(\phi_1-\lambda_jI)^px = 0\}.$
But even if we don't have that, letting $V_{\lambda_j}$ be as above, we still get the obvious inclusions:
Let $(a_1,\ldots,a_n)$ be a tuple corresponding to a nonzero $V_{(a_1,\ldots,a_n)}$ from the first decomposition. Then each $a_i$ must be an eigenvalue of $\phi_i$. In particular, $a_1 = \lambda_j$ for some $j$, $1\leq j\leq k$. Then $V_{(a_1,\ldots,a_n)} = V_{(\lambda_j,a_2,\ldots,a_n)} \subseteq V_{\lambda_j}.$
In general, if we let $V_{j,\mu} = \{x\in V\mid \text{there exists }p\text{ such that }(\phi_j-\mu I)^px=0\},$ then $V_{(a_1,\ldots,a_n)}\subseteq V_{1,a_1}\cap V_{2,a_2}\cap\cdots\cap V_{n,a_n}.$ This inclusion holds in general, whether or not the characteristic polynomials of the $\phi_i$ split.