Let me sketch a proof of existence of the Jordan canonical form which, I believe, makes it somewhat natural.
Let us say that a linear endomorphism $f:V\to V$ of a nonzero finite dimensional vector space is decomposable if there exist proper subspaces $U_1$, $U_2$ of $V$ such that $V=U_1\oplus U_2$, $f(U_1)\subseteq U_1$ and $f(U_2)\subseteq U_2$, and let us say that $f$ is indecomposable if it is not decomposable. In terms of bases and matrices, it is easy to see that the map $f$ is decomposable iff there exists a basis of $V$ such that the matrix of $f$ with respect to which has a non-trivial diagonal block decomposition (that it, it is block diagonal two blocks)
Now it is not hard to prove the following:
Lemma 1. If $f:V\to V$ is an endomorphism of a nonzero finite dimensional vector space, then there exist $n\geq1$ and nonzero subspaces $U_1$, $\dots$, $U_n$ of $V$ such that $V=\bigoplus_{i=1}^nU_i$, $f(U_i)\subseteq U_i$ for all $i\in\{1,\dots,n\}$ and for each such $i$ the restriction $f|_{U_i}:U_i\to U_i$ is indecomposable.
Indeed, you can more or less imitate the usual argument that shows that every natural number larger than one is a product of prime numbers.
This lemma allows us to reduce the study of linear maps to the study of indecomposable linear maps. So we should start by trying to see how an indecomposable endomorphism looks like.
There is a general fact that comes useful at times:
Lemma. If $h:V\to V$ is an endomorphism of a finite dimensional vector space, then there exists an $m\geq1$ such that $V=\ker h^m\oplus\def\im{\operatorname{im}}\im h^m$.
I'll leave its proof as a pleasant exercise.
So let us fix an indecomposable endomorphism $f:V\to V$ of a nonzero finite dimensional vector space. As $k$ is algebraically closed, there is a nonzero $v\in V$ and a scalar $\lambda\in k$ such that $f(v)=\lambda v$. Consider the map $h=f-\lambda\mathrm{Id}:V\to V$: we can apply the lemma to $h$, and we conclude that $V=\ker h^m\oplus\def\im{\operatorname{im}}\im h^m$ for some $m\geq1$. moreover, it is very easy to check that $f(\ker h^m)\subseteq\ker h^m$ and that $f(\im h^m)\subseteq\im h^m$. Since we are supposing that $f$ is indecomposable, one of $\ker h^m$ or $\im h^m$ must be the whole of $V$. As $v$ is in the kernel of $h$, so it is also in the kernel of $h^m$, so it is not in $\im h^m$, and we see that $\ker h^m=V$.
This means, precisely, that $h^m:V\to V$ is the zero map, and we see that $h$ is nilpotent. Suppose its nilpotency index is $k\geq1$, and let $w\in V$ be a vector such that $h^{k-1}(w)\neq0=h^k(w)$.
Lemma. The set $\mathcal B=\{w,h(w),h^2(w),\dots,h^{k-1}(w)\}$ is a basis of $V$.
This is again a nice exercise.
Now you should be able to check easily that the matrix of $f$ with respect to the basis $\mathcal B$ of $V$ is a Jordan block.
In this way we conclude that every indecomposable endomorphism of a nonzero finite dimensional vector space has, in an appropriate basis, a Jordan block as a matrix.
According to Lemma 1, then, every endomorphism of a nonzero finite dimensional vector space has, in an appropriate basis, a block diagonal matrix with Jordan blocks.
Best Answer
Your answer is correct and your reasoning is mostly correct. However I don't understand how you can conclude that $f(a), b$ form a basis for a $2 \times 2$ Jordan block? For starters they have different eigenvalues so they can't possibly be in the same Jordan block. Want you want to say is the following.
There are at least two distinct $0$-Jordan blocks, corresponding to $\operatorname{ker}(f)$, and $\operatorname{ker}(f^{2})$. From the data you have got it you should be able to see that $\operatorname{dim}\operatorname{ker}(f) = 2$, and that $$ \operatorname{dim}\operatorname{ker}(f^{2}) > \operatorname{dim}\operatorname{ker}(f) = 2. $$ But since $$ \operatorname{dim}\operatorname{ker}(f - \operatorname{Id}) \geq 1, $$ we have that $$ 3 \leq \operatorname{dim}\operatorname{ker}(f^{2}) \leq 4 - 1 = 3, $$ and so $\operatorname{dim}\operatorname{ker}(f^{2}) = 3$. Thus we have
$$ \operatorname{dim}\operatorname{ker}(f) = 2, \ \operatorname{dim}\operatorname{ker}(f^{2}) = 3, \ \operatorname{dim}\operatorname{ker}(f - \operatorname{Id}) = 1 $$
Then our Jordan decomposition is
$$ \operatorname{ker}(f) \oplus \left(\operatorname{ker}(f^{2}) / \operatorname{ker}(f)\right) \oplus \operatorname{ker}(f - \operatorname{Id}) $$
With $\operatorname{dim}\operatorname{ker}(f) = 2, \operatorname{dim}\left( \operatorname{ker}(f^{2}) / \operatorname{ker}(f) \right) = 1$, and $\operatorname{dim}\operatorname{ker}(f - \operatorname{Id}) = 1$.
Thus we have a $(2\times 2)$-Jordan block for eigenvalue zero, a $(1 \times 1)$-Jordan block for eigenvalue zero, and a $(1 \times 1)$-Jordan block for eigenvalue $1$. So we are done.
In case there is any confusion, the bases for the Jordan blocks are:
$(2 \times 2)$-$0$-Jordan block: $\{f(a), a\}$,
$(1 \times 1)$-$0$-Jordan block: $\{u \}$ where $u \in \ker{f} \backslash \operatorname{Span}\{ f(a) \}$,
$(1 \times 1)$-$1$-Jordan block: $\{ b \}$