Push Forward on product manifold. Some words before the question.
For two smooth manifolds $M$ and $P$ It is true that
$T(M\times P)\simeq TM\times TP $
If I have local coordinates $\lambda$ on $M$ and $q$ on $P$ then ($\lambda$, $q$) are local coordinates on  $M\times P$ (right?). This means that in these local coordinates the tanget vectors are of the form $a^{i}\frac{\partial}{\partial\lambda^{i}}+b^{i}\frac{\partial}{\partial q^{i}}$
Now, I can compute push forwards in local coordinates. For example, for a function
$f(\lambda, q)\rightarrow(\lambda,Q(q,\lambda))$
Then
$f^{*}\left(\frac{\partial}{\partial\lambda}\right)=\frac{\partial}{\partial\lambda}+\frac{\partial Q}{\partial\lambda}\frac{\partial}{\partial q}$
where I just had to do the matrix product of the Jacobian to the column vector $(1,0)^{T}$.
Actual Question.
For a function $f:\, M\times P\longrightarrow M\times P$
and without using local coordinates what can be said about the Push forward $f^{*}:\, TM\times TP\longrightarrow TM\times TP$       ?.
Particularly interested if the push forward can be decomposed into something in $TM$ plus something in $TP$
 A: Not entirely sure if this is what you're asking for or not, but let's give this a go.
Observe that at any point $(p,q)\in P\times Q$ we can write 
$$T_{(p,q)}P\times Q\simeq T_pP\oplus T_qQ,$$
where $T_qQ$ is imbedded in $T_{(p,q)}P\times Q$ as the tangent space of the submanifold $\{p\}\times Q$, and similarly for $T_pP$.
Then if $\phi :P\times Q\to P\times Q$ is a smooth map, and $\phi(p,q)=(p',q')$, its differential gives a map
$$\phi_* : T_pP\oplus T_qQ \to T_{p'}P\oplus T_{q'}Q.$$
Therefore we can write $\phi_*$ as a matrix 
$$\phi_* = \begin{pmatrix} A & B \\ C & D \end{pmatrix},$$
where 
$$A : T_pP\to T_{p'}P,$$
$$B:T_qQ\to T_{p'}P,$$
$$C:T_pP\to T_{q'}Q,$$
$$D:T_qQ\to T_{q'}Q.$$
You can check that if we unravel the definitions, we have $A$ is the differential of the map $P\to P'$ defined by $p\mapsto \pi_1(\phi(p,q))$, i.e., holding $q$ constant. Similarly, $B$ is the differential of the map $Q\to P$ given by $q\mapsto \pi_1(\phi(p,q))$.
We get similar results for $C$ and $D$.
Let's not focus too much on the details, since I think this will become more clear if we apply it to your particular case of interest.
Let $G$ and $H$ be Lie groups, $\alpha : G\to H$ a Lie group homomorphism.
Define $\phi : G\times H \to G\times H$ by $(g,h)\mapsto (g,\alpha(g)h)$. We want to compute its differential.
$A$ is the differential of the map $g\mapsto g$, since we're holding $h$ constant and projecting onto the first factor. Thus $A=1$ is the identity map.
$B$ is the differential of the map $h\mapsto g$, holding $g$ constant. Thus $B=0$.
$C$ is the differential of the map $g\mapsto \alpha(g)h$, holding $h$ constant, so $C=r_{h*}\alpha_*$.
Finally $D$ is the differential of the map $h\mapsto \alpha(g)h$, holding $g$ constant, so $D=\ell_{\alpha(g)*}$.
Thus we have that the differential of $(g,h)\mapsto (g,\alpha(g)h)$ is given at a point $(g,h)$ by
$$\begin{pmatrix} 1 & 0 \\ r_{h*}\alpha_* & \ell_{\alpha(g)*} \end{pmatrix}.$$
In the particular case where $G\subseteq H$, so $\alpha$ is just the inclusion, we have that the differential is
$$\begin{pmatrix} 1 & 0 \\ r_{h*} & \ell_{g*} \end{pmatrix}.$$
If $X\in \mathfrak{g}\subseteq \mathfrak{h}$, $Y\in\mathfrak{h}$, then in this last case, we have 
$$\phi_*(X,Y) = \begin{pmatrix} 1 & 0 \\ r_{h*} & \ell_{g*} \end{pmatrix}\begin{pmatrix} X \\ Y\end{pmatrix} = \begin{pmatrix} X \\ Xh + gY \end{pmatrix}$$
If we're identifying the tangent spaces via left multiplication, then $gY_h\in T_{gh}H$ corresponds to the same element of $\mathfrak{h}$ as $Y_h$, and $X_gh=gX_eh = ghh^{-1}X_eh = gh\operatorname{Ad}_{h^{-1}}X_e$.
Thus we end up with $\phi_*(X,Y) = (X,\operatorname{Ad}_{h^{-1}}X+Y)$.
