It is better to see the connection on the level of semiflows generated by the equations. Namely, let $\varphi^{t}$ be the semiflow generated by $(1)$ and $\psi^{t}$ be the semiflow generated by $(2)$, i.e. $\varphi^{t}(u_{0})=u(t;u_{0})$ and $\psi^{t}(v_{0})=v(t;v_{0})$, where $u(t;u_{0})$ solves $(1)$ such that $u(0,u_{0})=u_{0}$ and $v(t;v_{0})$ solves $(2)$ such that $v(0;v_{0})=v_{0}$. Then we have
$$ \varphi^{t}(u_{0}) = \psi^{\int_{0}^{t}\gamma(\varphi^{s}(u_{0}))ds}(u_{0}).$$
We need to calculate the differential $d_{u_{0}}$ at $u_{0}$. Using the chain rule, we have
$$ d_{u_{0}}\varphi^{t} = \left.\frac{d}{d\tau}\psi^{\tau}(u_{0})\right|_{\tau=\int_{0}^{t}\gamma(\varphi^{s}(u_{0}))ds} \circ\left(\int_{0}^{t} d_{\varphi^{s}(u_{0})}\gamma \circ d_{u_{0}}\varphi^{s} ds\right) + \left.d_{u_{0}}\psi^{\tau}\right|_{\tau=\int_{0}^{t}\gamma(\varphi^{s}(u_{0}))ds}.$$
Now note that
$$\left.\frac{d}{d\tau}\psi^{\tau}(u_{0})\right|_{\tau=\int_{0}^{t}\gamma(\varphi^{s}(u_{0}))ds(u_{0})} = F(\psi^{\int_{0}^{t}\gamma(\varphi^{s}(u_{0}))ds}(u_{0})) = F(\varphi^{t}(u_{0})).$$
In the case $u_{0}$ is a $T$-periodic orbit of $\varphi$, we put $t=T$ in the above equations and obtain
$$ d_{u_{0}}\varphi^{t} = F(u_{0}) \circ \left(\int_{0}^{T} d_{\varphi^{s}(u_{0})}\gamma \circ d_{u_{0}}\varphi^{s} ds\right) + d_{u_{0}}\psi^{\tau(T)},$$
where $\tau(T) = \int_{0}^{T}\gamma(\varphi^{s}(u_{0}))ds$. Note that $d_{u_{0}}\varphi^{T}$ is the monodromy operator of $(1)$ and $d_{u_{0}}\psi^{\tau(T)}$ is the monodromy operator of $(2)$ along $u_{0}$. Put (the product of $n$-column (left) and $n$-row (right))
$$ W:= F(u_{0}) \circ \left(\int_{0}^{T} d_{\varphi^{s}(u_{0})}\gamma \circ d_{u_{0}}\varphi^{s} ds\right).$$
Note that $W$ is a rank-1 operator with the image spanned by $F(u_{0})$. More precisely, for each column-vector $\xi \in \mathbb{R}^{n}$ we have
$$W\xi = \Lambda(\xi) F(u_{0}), \text{ where } \Lambda(\xi) = \left(\int_{0}^{T} d_{\varphi^{s}(u_{0})}\gamma \circ d_{u_{0}}\varphi^{s} ds\right)\xi.$$
Thus, $d_{u_{0}}\varphi^{T}$ is given by the rank-1 perturbation of $d_{u_{0}}\psi^{\tau(T)}$ by $W$. However, it is very specific. Namely, put $\xi_{0} := F(u_{0})$. Clearly, $(d_{u_{0}}\psi^{\tau(T)}) \xi_{0} = \xi_{0}$ and $(d_{u_{0}}\varphi^{T})\xi_{0} = \xi_{0}$ since the velocity vector corresponds to the unit Floquet multiplier. On the other hand,
$$\xi_{0} = (d_{u_{0}}\varphi^{T})\xi_{0} = (W + d_{u_{0}}\psi^{\tau(T)})\xi_{0} = (\Lambda(\xi_{0})+1)\xi_{0}.$$
Thus, $\Lambda(\xi_{0}) = 0$.
Suppose that $\xi_{*}$ is a nontrivial Floquet vector of $(2)$ for some multiplier $\mu \in \mathbb{C}$, i.e. $d_{u_{0}}\psi^{\tau(T)} \xi_{*} = \mu \xi_{*}$. Consider $\xi := \alpha \xi_{0} + \beta \xi_{*}$ for some $\alpha$ and $\beta \not=0$. Then
$$ (d_{u_{0}}\varphi^{t}) \xi = \beta\Lambda(\xi_{*})\xi_{0} + \alpha\xi_{0} + \beta \mu \xi_{*}.$$
Clearly, $(d_{u_{0}}\varphi^{t}) \xi = \kappa \xi$ if and only if $\kappa = \mu$ and
$$\beta\Lambda(\xi_{*}) + \alpha = \alpha \mu.$$
In the case $\Lambda(\xi_{*}) \not= 0$, we put $\alpha = 1$ and $\beta = (\mu - 1)/\Lambda(\xi_{*})$. In the case $\Lambda(\xi_{*}) = 0$ we put $\alpha = 0$ and $\beta = 1$. Then we get that $\xi$ is a Floquet vector of $(1)$ corresponding to the Floquet multiplier $\mu$.
Consequently, the Floquet multipliers of $(1)$ and $(2)$ are the same. It is only the Floquet vectors that are affected (in the direction of the velocity vector $F(u_{0})$) by the time scaling.
