I am just learning about the concept of transversality so please be patient with me. So far I understand that a map $f$ is transversal to a submanifold $Z$ if $\text{Im}(d_xf) + T_yZ = T_yY$. Now I consider this example: $f:\mathbb{R} \to \mathbb{R}^2$, $x \mapsto (0,x)$ and $Z$ is the x-axis.
Now I try to construct the tangent spaces (also this is kind of new to me). The tangent space of $f$ should be the map $(0,v)$ for $v \in \mathbb{R}$. But now I am stuck. The tangent space $T_yY$ should be $\mathbb{R}^2$ again, right? But as $Z$ is the x-axis isn't the tangent space for example at point $0$ just the y-axis? So I don't see how I get in the end $\mathbb{R}^2$. But maybe I made some mistakes in between?
Since $Y= \mathbb R^2$, you are correct that $T_y Y = \mathbb R^2$ for any $y\in \mathbb R^2$. As $Z$ is the $x$-axis, the tangent space contains all those vectors that are tangent to $Z$, thus for all $y=(x, 0) \in Z$, $T_yZ$ is also the $x$-axis.
You are sort of correct concerning the "tangent space of $f$. Indeed you need to find $\text{Im} (d_y f)$ instead of the tangent space of $\mathbb R$. For each $t\in \mathbb R$, the tangent map $d_t f : \mathbb R^2 \to T_{(f(t)} Y \cong \mathbb R^2$ is given by $v\mapsto (0,v)$. As a result, since $\text{Im} (f) \cap Z = \{y= (0,0)\}$ and at this point we have
$$ \text{Im} d_y f + T_y Z = \mathbb R^2,$$
thus $f$ is transversal to $Z$.
As an example, consider the mapping $g (t) = (0,t^3)$. Although this map has the same image as $f$ (that is, $f(\mathbb R) = g(\mathbb R)$), $g$ is not transvarsal to $Z$ since at $y = (0,0)$, the tangent map $d_0 g$ is zero.