Triangular Matrices and Simple Modules 
Let $\Bbb{T}_n(k)=\{n \times n \text{ upper triangular matrices (which includes the diagonal entries)}\}$
I want to first express $\Bbb{T}_{n}(k)$ (as a $\Bbb{T}(k)$-module) as a direct sum of projective and indecomposable modules. Then I want to write the Jordan-Hölder decomposition for each of those modules.

Let $Col(i) =$ { n x n matrices with zeros everywhere except for the ith column; in the ith column, we have zeros below the diagonal entry and possibly nonzero entries in the diagonal and above the diagonal}. So, for example, a $3 \times 3$ matrix in $ Col(2)$ looks like $\begin{bmatrix} 0 & a & 0 \\0 & b & 0 \\0 & 0 & 0 \end{bmatrix}$ for some $a,b$.
Obviously, for any $A \in \Bbb{T}_n(k)$, we can write $A$ as a sum of the elements from $Col(1),...,Col(n)$. So, for example,
$\begin{bmatrix} a & b & c \\0 & d & e \\0 & 0 & f \end{bmatrix} = \begin{bmatrix} a & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & b & 0 \\0 & d & 0 \\0 & 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 & c \\0 & 0 & e \\0 & 0 & f \end{bmatrix}.$
Obviously $Col(i) \cap (\sum_{i \not= j} Col(j)) = \{0\}$.
So $\Bbb{T}_n(k) = Col(1) \oplus ... \oplus Col(n)$.
Every $Col(i)$ is free, where the basis element is the matrix with zeros everywhere except for the ith diagonal entry where we have 1. So, for example, for any random element $\begin{bmatrix} 0 & a & 0 \\0 & b & 0 \\0 & 0 & 0 \end{bmatrix} \in Col(2)$, we get $$\begin{bmatrix} 0 & a & 0 \\0 & b & 0 \\0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & a & 0 \\0 & b & 0 \\0 & 0 & 0 \end{bmatrix}$$
But free modules are projective, so each $Col(i)$ must be projective. They are also indecompomsable. If $Col(i) = P \oplus Q$, then $P \cap Q = \{0\}$. But then the basis element is either contained in $P$ or in $Q$. But the summand that contains the basis element contains all of $Col(i)$, so the other summand must be zero. So $Col(i)$ is indecomposable.
Now I have to write the Jordan-Holder decomposition for $Col(i)$, and this is where I started getting confused. For example, if we look at $Col(3)$ for the $3 \times 3$ matrices...we know that we have a submodule of $Col(3)$, which is the set of all matrices with zeros everywhere except for the first two rows of the third column. And a submodule of that is the set of all matrices with zeros everywhere except for the first row of the third column. So...
$ \langle \begin{bmatrix} 0 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 1 \end{bmatrix} \rangle \supsetneqq \langle \begin{bmatrix} 0 & 0 & 0 \\0 & 0 & 1 \\0 & 0 & 0 \end{bmatrix} \rangle \supsetneqq \langle \begin{bmatrix} 0 & 0 & 1 \\0 & 0 & 0 \\0 & 0 & 0 \end{bmatrix} \rangle \supsetneqq 0$.
The problem here is that the quotients are not modules. For instance if we look at $\langle \begin{bmatrix} 0 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & 1 \end{bmatrix} \rangle / \langle \begin{bmatrix} 0 & 0 & 0 \\0 & 0 & 1 \\0 & 0 & 0 \end{bmatrix} \rangle$, we get matrices with zeros everywhere except for the entry in the 3rd row and the 3rd column...but this set of matrices is not closed under scalar multiplication. So I must be doing something wrong, but I can't really see it...
Thanks in advance
 A: Let $R:=\Bbb T_n(k)$.
Problem spot 1

Every $Col(i)$ is free

No, while they are certainly free $k$ modules, they are not free $R$ modules. For one thing $R$ is $n(n+1)/2$ dimensional, while one of those column left ideals is never more than $n$ dimensional, so it's impossible for a left ideal to be isomorphic to copies of $R$.
However, as you noted, the $Col(i)$ are all direct summands of $R$, and so they are projective modules by virtue of being direct summands of the free module $_RR$.
Problem spot 2

They are also indecompomsable. If Col(i)=P⊕Q, then P∩Q={0}. But then the basis element is either contained in P or in Q. But the summand that contains the basis element contains all of Col(i), so the other summand must be zero. So Col(i) is indecomposable.

I have a hard time following this, and it seems like it is still based on the misapprehension that the column ideals are free. (Added later: We discovered in the comments that the root of the problem is that Artus was using the basis definition of "free module," but did not notice that the basis elements s/he was looking at were not $R$ independent. Mini problem spot: Even if $P\oplus Q\cong R$, there is no reason to expect that a basis element of $P\oplus Q$ must be contained in $P$ or contained in $Q$, as you tried above. In fact, if $P$ and $Q$ are nonzero, that will never happen. If the generator were in $P$, its span would be trapped in $P$. How would it generate all of $P\oplus Q$?)
It is very easy to see that the column ideals are irreducible if you first prove this useful lemma, which I should leave up to you:
Lemma

The submodules of $Col(i)$ are linearly ordered, and they are exactly what you expect them to be: you get them by turning zeroing out coordinates in the column one at a time from the bottom up.

For example, the left ideals in $\begin{bmatrix}0&0&k\\0&0&k\\0&0&k\end{bmatrix}$ are $\begin{bmatrix}0&0&k\\0&0&k\\0&0&0\end{bmatrix}$ and $\begin{bmatrix}0&0&k\\0&0&0\\0&0&0\end{bmatrix}$.
Using this, it's obvious $Col(i)$ is indecomposable: the intersection of two nonzero submodules is always the smaller submodule in the chain of submodules.
This lemma also automatically gives you the Jordan-Holder decomposition! When the submodules are in a chain, the decomposition is unique! To show that the factor modules are indeed simple, just note that their $k$ dimension is $1$, and you can't get a nonzero module which is smaller.

Your composition series is perfect!
Problem spot 3
You're misidentifying the quotient with a subset of R. It isn't a set of matrices, it's a set of cosets of the left ideal. You're right that you can pick a set of coset representatives that is nonzero only on the 3-3 entry, but those aren't the elements of the quotient. Coset representatives are rarely closed under addition. The cosets are closed under addition and left multiplication by R though.
