Question: Both the null space and the eigenspace are defined to be "the set of all eigenvectors and the zero vector". They have the same definition and are thus the same. Is there ever a scenario where the null space is not the same as the eigenspace (i.e., there is at least one vector in one but not in the other)?

Answer: 
Technically it doesn't.

If $f=0$ then $\mathcal{N}(f) = \mathbb{R}^2$. But otherwise, the image of $f$ is of dimension at least $1$. Moreover since the image of $f$ is a subspace of $\mathbb{R}$ the dimension of the image is also at most $1$. Thus, by the rank-nullity theorem we have that $\mathcal{N}(f)$ is a vector space of dimension $2-1=1$. 

Any vector space of dimension $1$ on $\mathbb{R}^2$ takes the form $\text{span}(v)$ where $v\not =0$ is some vector of $\mathbb{R}^2$. Hence, is a line which passes through the origin in the direction of $v$.

One last thing. The word "bounded" is irrelevant here. All linear functionals from $\mathbb{R}^2$ to $\mathbb{R}$ are bounded.
A plain, old-fashioned, vanilla-flavored, very basic demonstration which doesn't even need the rank-nullity theorem, though that is surely a *good* theorem:

Let us assume that

$f \ne 0; \tag 1$

for otherwise, we clearly have

$\mathcal N(f) = \Bbb R^2, \tag 2$

the trivial case. Now for *any* such linear $f$, trivial or not, $(x, y) \in \mathcal N(f)$ when

$f(x, y) = f(x(1, 0) + y(0, 1)) = xf(1, 0) + yf(0, 1) = 0; \tag 3$

now if $f \ne 0$, at least one of

$f(1, 0), f(0, 1) \ne 0; \tag 4$

suppose then that

$f(1, 0) \ne 0; \tag 5$

thus,

$x = -\dfrac{f(0, 1)}{f(1, 0)} y; \tag 6$

this clearly describes a line through the origin; likewise if

$f(0, 1) \ne 0 \tag 7$

then

$y = -\dfrac{f(1, 0)}{f(0, 1)} x;, \tag 8$

also the equation of a line through $(0, 0)$.

***Note Added in Edit, Wednesday 27 February 2019 10:16 AM PST:*** The above easily generalizes to show that for linear

$f:\Bbb R^n \to \Bbb R, \tag 9$

$\mathcal N(f)$ is a hyperplane containing $0 \in \Bbb R^n$; for if we adopt the standard basis

$\mathbf e\_i \in \Bbb R^n, \; 1 \le i \le n, \tag{10}$

then

$f \left ( \displaystyle \sum\_1^n x\_i \mathbf e\_i \right ) = \displaystyle \sum\_1^n x\_i f(\mathbf e\_i); \tag{11}$

if $f \ne 0$ then at least one of

$f(\mathbf e\_i) \ne 0; \tag{12}$

if

$f(\mathbf e\_j) \ne 0, \tag{13}$

we have *via* (11),

$f \left ( \displaystyle \sum\_1^n x\_i \mathbf e\_i \right ) = 0 \Longrightarrow x\_j = -\displaystyle \sum\_{j \ne i = 1}^n \dfrac{f(\mathbf e\_i)}{f(\mathbf e\_j)} x\_i, \tag{14}$

the equation of a hyperplane containing the origin. ***End of Note.***