One can test one's understanding of this question on the matrix A=[0,1|0,0], since this example already exhibits many of the relevant features of the problem. For eample, one cannot find two linearly independent vectors v and w such that Av=Aw=0 (try to solve for v the equation Av=0) although 0 is a double root of the characteritic polynomial det(XI-A)=X^2.
This is not specific to the eigenvalue 0. Recall that for any eigenvalue one distinguishes its algebraic multiplicity n (in this case, n=2) from its geometric multiplicity m (in this case, m=1). See here for more details and a worked out example.
If $A$ is $2 \times 2$, has eigenvalue $\lambda=1$ with algebraic multiplicity $2$ and geometric multiplicity $1$, then the null space (i.e. kernel) of $A-1I_2$ is 1-dimensional. In your example, it's the span of $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. The null space of $(A-1I_2)^2$ is 2-dimensional (and is all of $\mathbb{F}^2$ where $\mathbb{F}$ is whatever field of scalars you're working over -- probably $\mathbb{R}$ or $\mathbb{C}$). Notice that $(A-1I_2)\begin{bmatrix} a\\ b\end{bmatrix} = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} a\\ b \end{bmatrix} = \begin{bmatrix} b \\ 0 \end{bmatrix}$. So as long as $b=1$, you get $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$. Thus $M=\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}$ for any $a \in \mathbb{F}$ satisfies $M^{-1}AM=J$. So for question #1, "Yes" any scalar will do. This is analogous to the fact that when a matrix is diagonalizable it is diagonalized by infinitely many different matrices. For example: if $P^{-1}BP=D$ is diagonal, then replacing any column in $P$ by a non-zero multiple will also result in a matrix such that $Q^{-1}BQ=D$.
For question #2. The answer in general is "No". In your case and many "small" examples this works out. But in general if you have several Jordan blocks involving the same eigenvalue, this will fail.
Let me use an example to demonstrate. Let $C = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$. The only eigenvalue is $\lambda=2$. It's algebraic multiplicity is 3 and geometric multiplicity is 2 [Note: The geometric multiplicity of an eigenvalue is equal to the number of Jordan blocks associated with that eigenvalue --- what you are calling Jordan miniblocks are actually the Jordan blocks --- each new Jordan block gives you one new (linearly independent) eigenvector.] Also, $C$ is already in Jordan form. It consists of 2 Jordan blocks: $J_1 = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$ and $J_2 = [2]$.
The null space of $C-2I_3$ is 2-dimensional: $C-2I_3=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$. So the null space of $C-2I_3$ is spanned by $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$.
The null space of $(C-2I_3)^2$ is 3-dimensional: $(C-2I_3)^2=0$ (this null space is all of $\mathbb{F}^3$). In summary, we have 2 (linearly independent) eigenvectors with eigenvalue 2 and 1 generalized eigenvector (which is not a regular eigenvector).
Suppose for some odd reason we came up with ${\bf v}_1 = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}$ and ${\bf v}_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$ as our basis of eigenvectors (this is indeed a basis -- although definitely not the "natural" choice). If we try to solve $(A-2I_3){\bf x} = {\bf v}_1$ or $(A-2I_3){\bf x} = {\bf v}_2$ to find our generalized eigenvector, we get stuck! Both of these systems are inconsistent: $\begin{bmatrix} 0 & 1 & 0 & : & 1 \\ 0 & 0 & 0 & : & 0 \\ 0 & 0 & 0 & : & 1 \end{bmatrix}$ and $\begin{bmatrix} 0 & 1 & 0 & : & 0 \\ 0 & 0 & 0 & : & 0 \\ 0 & 0 & 0 & : & 1 \end{bmatrix}$
This doesn't mean that $(A-2I_3)^2{\bf x} = {\bf 0}$ has no new solutions beyond ${\bf v}_1$ and ${\bf v}_2$. But it does mean that any new solutions cannot be obtained by "backing-up" from ${\bf v}_1$ or ${\bf v}_2$.
In general you cannot find a basis which puts your matrix into Jordan form by finding a basis of eigenvectors and "backing-up" chains.
To find your desired basis you must start with the longest chain, find the top of that chain, then find the top of the next longest chain (excluding what you just found) and continue working to the shortest chain.
In my example, I would find a basis for $\mathrm{Null}((A-2I_3)^1)$, say $\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. Then complete it to a basis for $\mathrm{Null}((A-2I_3)^2)$. The new vector that showed up say ${\bf w}_1 = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$ would then belong to $\mathrm{Null}((A-2I_3)^2)$ but not to $\mathrm{Null}((A-2I_3)^1)$. So ${\bf w}_1$ is a generalized eigenvector but not an eigenvector and since it lives in $\mathrm{Null}((A-2I_3)^2)$ it is the start of a 2-chain. Let ${\bf w}_2 = (A-2I_3){\bf w}_1$ to complete this chain. I could then take ${\bf w}_2$ and complete it to a basis for $\mathrm{Null}((A-2I_3)^1)$. Call the new basis vector ${\bf w}_3$. Then $M = [{\bf w}_2 \; {\bf w}_1 \; {\bf w}_3]$ will put $C$ into its Jordan form [I reversed the order of my chain to make the $1$'s appear above (instead of below) the diagonal in the Jordan form].
A general procedure for finding the Jordan form works in a similar manner. Since it involves a lot of finding and completing bases, there can be a lot of variability in what matrix $M$ you end up with (as brought up in question #1).
Question #2b "Does a basis always exist which puts a matrix into Jordan form?" "Yes" if the characteristic polynomial factors into linear factors over your field of scalars (so always yes over the complex numbers). Why? Now that's complicated. I recommend "Spence, Insel, and Friedberg's Linear Algebra" for a reasonably easy readable treatment.
Best Answer
I will assume that you are capable of finding the eigenvector for the simple root, and only focus on the repeated eigenvalue $x_1=1$. Let $\mathbf{v}=\begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}$ and consider $(A-I)\mathbf{v}=\mathbf{0}$. This yields $\begin{bmatrix} 0 \\ 0 \\ 8a+6b+2c \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$ in which case $8a+6b+2c=0$. Let $b$ and $c$ be free variables, giving us $\mathbf{v}=\begin{bmatrix} -\frac{3}{4}b-\frac{1}{4}c \\ b \\ c \\ \end{bmatrix}=b\begin{bmatrix} -\frac{3}{4} \\ 1 \\ 0 \\ \end{bmatrix}+c\begin{bmatrix} -\frac{1}{4} \\ 0 \\ 1 \\ \end{bmatrix}$.
I hope this helps.