Is a linear transformation from a higher dimension to a lower dimension always onto? Is a linear transformation from a higher dimension to a lower dimension always onto?
I also heard that for a transformation to be onto, the transformation matrix $A$, say $m \times n$, has to have $m$ pivot columns.
So does that mean $2 \times 3$ matrix with $1$'s in every spot is not able to map a vector from $\mathbb R^3$ to the span of $\mathbb R^2$?
 A: No, it is not.
The simplest example would be the null map, which sends everyone to $0$. It is linear, but almost never onto.
Other examples would be the projections on a sub-vector space:
Let $E$ and $F$ be two vector spaces of finite dimensions $m$ and $n$ respectively (with $m>n$.
Then take $H$ a sub-vector space of $F$ of dimension $k<n$.
Define $p_H$ the projection on $H$.
Then $p_H$ is linear, but not onto because $\mathrm{dim}(\mathrm{Im}(p_H))=\mathrm {dim}(H)=k<n$.

In your particular example, the rank of your matrix is $1$, so it will span a space of dimension $1$, so it can be $\mathbb R^2$. So it is indeed not onto.
A: No, this is not the case. Take $T: \mathbb{R}^4 \to \mathbb{R}$ such that $T(v) = 1$ for all $v \in \mathbb{R}^4$. Then, $\text{im $T$} = \{1\} \neq \mathbb{R}$. 
Whether or not $T$ is onto depends on the nature of the mapping. We certainly could write a 
$T$ that maps $\mathbb{R}^4$ onto $\mathbb{R}$. For example, the canonical mapping would send $(x,0,0,0)$ to $x$. (The additionl details of the mapping are not relevant.) 
On the flip side, $T$ cannot be 1-to-1. This is the case from any mapping for a higher-dimensional space to a lower-dimensional space. It is easy to see that such a transformation has a non-trivial kernel and, hence, a non-full image. 
