When a system is controllable then it is indeed also 0-controllable, but as you stated the reverse does not have to hold. The term 0-controllable is related to stabilizability, namely the uncontrollable modes have to be stable. But for 0-controllable the state has to be able to reach zero in finite time, instead of eventually converge to zero for stabilizability. For continuous LTI systems every stable uncontrollable mode would die out to zero exponentially, so would take an infinite amount of time to actually reach zero. Therefore a continuous LTI system is only 0-controllable if it is controllable as well. For discrete LTI systems this is not completely the case, since when an uncomtroppable mode has an eigenvalue of zero, it will die out in finite amount of time.
In order to check whether a discrete LTI system is 0-controllable you either want a mode to be controllable or have an eigenvalue of zero, such any uncontrollable modes would die out in finite time. The easiest tool to check this in my opinion would be the Hautus lemma, namely for all $\lambda\neq0$ the following matrix should have full rank $[A-\lambda\,I \quad B]$.
As you have said, the uncontrollable subspace forms a complement to the controllable subpace. In other words, $\Bbb R^n = R \oplus I$. That is, we can uniquely decompose every vector $x \in \Bbb R^n$ into the form $x = x_R + x_I$, with the "controllable component" $x_R \in R$ and "uncontrollable component" $x_I \in I$. A state $x \in \Bbb R^n$ is controllable if and only if its uncontrollable component is zero.
It is useful to have such a decomposition because the nature of the "state-update" matrix $A$ is completely determined by its behavior over these separate subspaces, since for any $x = x_R + x_I$, we have
$$
Ax = A(x_R + x_I) = Ax_R + Ax_I.
$$
Here is a continuous-time example. Suppose that we have
$$
A = \pmatrix{a_1 & 0\\0 & a_2},\quad B = \pmatrix{1\\0}, \quad C = \pmatrix{1&1}, \quad D = 0.
$$
It is easy to verify that our controllable subspace of $\Bbb R^2$ is the $x_1$-axis, i.e. the span of $(1,0)$. Any other one-dimensional subspace can be selected as the uncontrollable subspace, but it is convenient to take $I$ to be the span of $(0,1)$, since this space happens to be invariant under $A$ (note: such a complement is not always available).
Suppose that the initial state is given by $x(0) = (x_1,x_2)$. It is easy to see that for input $u(t)$, the state and output will be
$$
x(t) = \left(x_1 + e^{a_1t}\int_0^t e^{-a_1t}u(t)\,dt, \quad x_2 e^{a_2t}\right),
\\
y(t) = \left[x_1 + e^{a_1t}\int_0^t e^{-a_1t}u(t)\,dt\right] + x_2 e^{a_2t}.
$$
The first component of the sum, which corresponds to the controllable component of $x(t)$, can be stabilized with a suitable input. The second component, which corresponds to the uncontrollable component of $x(t)$, cannot be stabilized in this way. We could also say that the component $x_2 e^{a_2}t$ is itself an autonomous trajectory of the system: it transpires independently of the input.
We see from the above that the output is only stabilizable (i.e. can be "steered" so that $y(t) \to 0$) if $e^{a_2t} \to 0$.
Correspondingly, we see that $a_2$ is an eigenvalue of $A$ whose eigenvector $(0,1)$ is an element of the uncontrollable subspace $I$.
Suppose that we keep $v_1 = (1,0)$ as the basis for $R$, but instead take $v_2 = (1,1)$ as a basis for $I$. We find that
$$
Av_1 = a_1 v_1 + 0v_2, \\
A v_2 = \pmatrix{a_1\\a_2} = (a_1 - a_2)v_1 + a_2 v_2.
$$
So, the matrix of $A$ relative to the basis $\{v_1,v_2\}$ is
$$
\bar A = \pmatrix{a_1 & a_1 - a_2\\0 & a_2}.
$$
We indeed find that the eigenvalue $a_2$ of $A$ is associated with our uncontrollable subspace $I$. It is tricky, however, to figure out exactly what "associated with $I$" really means here.
One way to make sense of it is this. If we define the projection map $P_I(x_R + x_I) = x_I$, then we could say that the eigenvalue $\lambda$ of $A$ is "associated with $I$" if it is an eigenvalue of the map $T:I \to I$ defined by $T(x) = P_I(Ax)$.
Best Answer
Actually, they are quite similar. When you are looking for stabilizability you don't care for the eigenvalues that are already stable. You can always left them "untouched" so you don't need them to be controllable for stabilization purposes. So, you only want controllability for the unstable eigenvalues, which is compatible with the first theorem.