Let $V$ be a symplectic space with symplectic form $(.,.)$. A transvection $\tau:V\rightarrow V$ is a linear map preserving $(.,.)$ such that it is identity on a hyperplane $W$ and also on $V/W$.
My aim is to understand how it looks?
We can write $V=\langle x\rangle \oplus W$. Then $\tau(x)=x+w_x$ for some $w_x\in W$ and $\tau$ is identity on $W$.
Let $v\in V$ be arbitrary. Write $v=ax+w$. Then $$\tau(v)=a\tau(x)+w=ax+aw_x+w=v+aw_x.$$
Question: How do we cover (determine) $a$ and $w_x$ in terms of $(.,.)$ and vector $v$?
The number $a$ and the vector $w_x$ do not depend solely on the symplectic form and on the transvection, but also on a choice you implicitly made.
The transvection produces a well-defined isomorphism $V \simeq V/W \oplus W$. You know that $V/W \simeq \mathbb{R}$, but this last isomorphism is not canonical in any way: there is no element in $V/W$ which could canonically be called $1$. Implicitly, you chose such an identification by your choice of $x \in V$, as $[x] \in V/W$ corresponds to $1 \in \mathbb{R}$.
As such, the expression $v = ax+w$ depends on this choice of identification, so $a$ depends on an arbitrary choice you made. For instance, I could have chosen instead $v = (\lambda a)(x/\lambda) + w$. Hence, there is no way to determine $a$ only from the symplectic form and the transvection.
Similarly, $w_x$ depends on the isomorphism $V/W \simeq \mathbb{R}$. Indeed, $$\tau(x/\lambda) = (1/\lambda)\tau(x) = (x/\lambda) + (w_x/\lambda) = (x/\lambda) + w_{x/\lambda}.$$
In a purely abstract formulation, the transvection $\tau$ determines and is determined by a linear map $$\tilde{\tau} : V/W \to W : [v=ax+w] = [ax] \mapsto aw_x.$$ Even though I expressed this map using an arbitrary isomorphism $V/W \simeq \mathbb{R} : [ax] \mapsto a$, it is defined independently of this choice as $aw_x = (\lambda a)w_{x/\lambda}$. By construction, we have $\tau(v) = v + \tilde{\tau}([v])$.