Let's investigate just two factors (extending this to arbitrarily finitely many isn't hard, but it takes more dots and more "index hell").
We want to show that given $V$, and two other vectors spaces $V_1,V_2$ along with four LINEAR maps:
$\pi_1:V \to V_1$
$\pi_2:V \to V_2$
$\epsilon_1:V_1 \to V$
$\epsilon_2:V_2 \to V$
such that:
$\pi_1\circ\epsilon_1 = 1_{V_1}$
$\pi_2\circ\epsilon_2 = 1_{V_2}$
and:
$\epsilon_1\circ \pi_1 + \epsilon_2\circ\pi_2 = 1_V$
that $V \cong V_1 \oplus V_2$.
So, one way we can do this is to exhibit an isomorphism:
$\phi: V \to V_1 \oplus V_2$ (Note I haven't said anything about whether or not the $V_i$ are even subspaces of $V$).
So here is how we will define $\phi$:
$\phi(v) = (\pi_1(v),\pi_2(v))$. Note this has the proper co-domain.
First, we'll show that $\phi$ is linear:
$\phi(v+v') = (\pi_1(v+v'),\pi_2(v+v')) = (\pi_1(v) + \pi_1(v'),\pi_2(v) + \pi_2(v'))$
$= (\pi_1(v),\pi_2(v)) + (\pi_1(v'),\pi_2(v')) = \phi(v) + \phi(v')$.
$\phi(av) = (\pi_1(av),\pi_2(av)) = (a\pi_1(v),a\pi_2(v)) = a(\pi_1(v),\pi_2(v)) = a\phi(v)$.
Now suppose that $\phi(v) = (0,0)$. We want to show $v = 0$.
Now $\phi(v) = (\pi_1(v),\pi_2(v))$, so if $\phi(v) = (0,0)$, we have (for this $v$):
$\pi_1(v) = 0$
$\pi_2(v) = 0$.
Since the $\epsilon_i$ are linear, we have: $\epsilon_1(0) = 0$, and $\epsilon_2(0) = 0$.
Thus $v = 1_V(v) = (\epsilon_1\circ \pi_1 + \epsilon_2\circ\pi_2)(v) = (\epsilon_1\circ \pi_1)(v) + (\epsilon_2\circ\pi_2)(v)$
$= \epsilon_1(\pi_1(v)) + \epsilon_2(\pi_2(v)) = \epsilon_1(0) + \epsilon_2(0) = 0 + 0 = 0$.
Thus $\phi$ is injective.
Before we tackle surjectivity, we prove a preliminary result:
$\pi_1 \circ \epsilon_2 = 0$
$\pi_2 \circ \epsilon_1 = 0$.
To prove this, we note that the $\pi_i$ are surjective, I will show this for $\pi_1$ (the proof is just the same for $\pi_2$):
Let $v_1 \in V_1$. Then we have $\pi_1(\epsilon_1(v_1)) = v_1$, so $v_1$ has the pre-image under $\pi_1$ of $\epsilon_1(v_1) \in V$.
Now for any $v \in V$, we have:
$v = \epsilon_1(\pi_1(v)) + \epsilon_2(\pi_2(v))$, so taking $\pi_2$ of both sides, we have:
$\pi_2(v) = \pi_2(\epsilon_1(\pi_1(v)) + \epsilon_2(\pi_2(v))) = \pi_2(\epsilon_1(\pi_1(v))) + \pi_2(\epsilon_2(\pi_2(v)))$
$= \pi_2(\epsilon_1(\pi_1(v))) + \pi_2(v)$.
Subtracting $\pi_2(v)$ from both sides, we have:
$0 = \pi_2(\epsilon_1(\pi_1(v)))$.
Now since $\pi_1$ is surjective, we can write ANY $v_1 \in V_1$ as $\pi_1(v)$ for SOME $v \in V$, so the above shows that (for such a $v$ so that $v_1 = \pi(v)$):
$0 = \pi_2(\epsilon_1(v_1))$, and thus this holds for any $v_1 \in V_1$.
The other condition is proved similarly.
Finally, let $(v_1,v_2)$ be any element of $V_1 \oplus V_2$.
Consider $\epsilon_1(v_1) + \epsilon_2(v_2) \in V$.
We have $\phi(\epsilon_1(v_1) + \epsilon_2(v_2)) = \phi(\epsilon_1(v_1)) + \phi(\epsilon_2(v_2))$
$= (\pi_1(\epsilon_1(v_1)),\pi_2(\epsilon_1(v_1))) + (\pi_1(\epsilon_2(v_2)),\pi_2(\epsilon_2(v_2)))$
$= (v_1,0) + (0,v_2) = (v_1,v_2)$, and $\phi$ is surjective.
