For Question 1,
The geometric interpretation largely has to do with the fact that if you have a vector $w$ that can be written as a linear combination of vectors $u$ and $v$ thus $w=u+v$ than if to see what happens to when we apply $T$ to $w$ (i.e. $T(w)$) we can think of this as what happens when we apply to $u$ and $v$ then just add those resulting vectors. Similarly the preservation of scalar multiplication is just saying we can see what happens when we apply $T$ to $u$ then scale it.
To really understand this, lets look at example. Let $T: \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ that reflects vector about y axis. Lets look at what this does would be for vector $(1,2)$, well first note that $(1,2)=(1,0)+2(0,1)$ thus when looking at $T[(1,2)]$ we can see that because property additivity and scalar multiplication that we have
$$T[(1,2)]=T[(1,0)]+2T[(0,1)]$$
Well when we reflect $(1,0)$ and $(0,1)$ about the y axis we get $(-1,0)$ and $(0,1)$ respectively thus we have
$$=(-1,0)+2(0,1)=(-1,2)$$
Thus we have $T[(1,2)]=(-1,2)$. Well what are we doing geometrically? Well what we did was we broke up $(1,2)$ addition of its x direction and y direction. We then scaled these down to make them unit vectors, we then flipped each one about the y axis, then rescaled them back, then added them back together to get our resulting vector $(-1,2)$.
Now as you may know that $(0,1)$ and $(1,0)$ is basis for vectors in $\mathbb{R}^{2}$. Well whenever we have have maps that have properties of linear maps this means we can always just find what the resulting vector is from applying a linear map to that vector, by just thinking of it as simply the method of scaling down vector to sum of basis vectors, transforming those basis vectors, then rescaling and adding these resulting vectors.
Im sorry if this answer isn't very good, and take it with a grain of salt, this is just how Ive always thought of it
For Question 2,
I don't think that this is necessarily true. For example what if you have map $T: \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ such that any vector it takes in it just transforms to 0 vector meaning for $v\in\mathbb{R}^{2}$ that
$$T(v)=0$$
well notice that
$$T(u+v)=0 \textrm{ and }T(v)+T(u)=0+0=0$$
along with for $c\in\mathbb{R}$
$$T(cv)=0 \textrm{ and }c(Tv)=c(0)=0$$
Thus $T$ is a linear map, but if you take the subspace $U=\{(a,a):a\in\mathbb{R}\}$ (i.e. line $x=y$) well this means for any $u\in U$ that $T(u)=0$ thus the image of U is just the 0 vector (i.e T(U)={0}) thus this linear map "transforms" a line into a point.
One consequence of the definition of a linear transformation is that every linear transformation must satisfy
$$ T(0_V)=0_W $$
where $0_V$ and $0_W$ are the zero vectors in $V$ and $W$, respectively. Therefore any function for which $T(0_V)\neq 0_W$ cannot be a linear transformation.
In your second example,
$$ T\Big(\begin{bmatrix}0\\0\end{bmatrix}\Big)=\begin{bmatrix}0\\1\end{bmatrix}\neq\begin{bmatrix}0\\0\end{bmatrix}$$
so this tells you right away that $T$ isn't linear.
Best Answer
Let $V$ be the vector space $\mathbb{R}$ over the field $\mathbb{Q}$. We take $T:V\to V$ given by $$T(x)=\begin{cases} 0 & x\in\mathbb{Q}\\x& \textrm{ else}\end{cases}$$
This satisfies $T(cv)=cT(v)$ for every $v\in V$ and every rational $c$. However $$T(1+\sqrt{2})+T(-\sqrt{2})\neq T(1)$$
Note that the first property implies $T(cv)=cT(v)$ for all $c\in \mathbb{N}$, and therefore all $c\in\mathbb{Q}$. Hence any function that has the first property but not the second must break the second at some irrational value.