There are two possible conventions for semidirect product, but let's suppose you're using the following one
$$
(g_1,h_2)\cdot(g_2,h_2) = (g_1g_2,h_1(\phi(g_1)h_2)).
$$
Employ the notation $\phi_g:H\to H, \ \phi_g(h) := \phi(g)h$ and $\phi^h:G\to H,\ \phi^h(g) := \phi(g)h$, and define
$$
\phi_g':= T_{e_H}\phi_g:\mathfrak{h}\to\mathfrak{h}, \qquad \dot{\phi}^h:=T_{e_G}\phi^h:\mathfrak{g}\to T_hH.
$$
So
$$
(g_1,h_1)\cdot (g_2,h_2)=(g_1g_2,h_1(\phi_{g_1}h_2))
\quad\textrm{and} \quad (g,h)^{-1} = (g^{-1},\phi_{g^{-1}}h^{-1}).
$$
Calculating $(g,h)\cdot(k,l)\cdot(g,h)^{-1}$, and differentiating wrt $(k,l)$, it is not difficult to show that the adjoint action of $G\ltimes H$ on $\mathfrak{g}\ltimes \mathfrak{h}$ is given by
$$
\operatorname{Ad}_{(g,h)}(\xi,\eta) = (\operatorname{Ad}_g\xi,\operatorname{Ad}_h(\phi'_g(\eta))+\sigma_h(\operatorname{Ad}_g\xi)),
$$
where
$$
\sigma_h:\mathfrak{g}\to\mathfrak{h}, \qquad \sigma_h(\xi):= h\cdot(\dot{\phi}^{h^{-1}}\xi).
$$
Here $\dot{\phi}^{h^{-1}}\xi\in T_{h^{-1}}H$, and $h\cdot $ denotes the derivative of left multiplication by $h$ (i.e., in general we define $h_1\cdot v_{h_2} := T_{h_2}L_{h_1}(v_{h_2})$, where $L_{h_1}:H\to H$ is left multiplication by $h_1$).
Now taking the derivative of this wrt $(g,h)$, we obtain an expression for the adjoint action of $\mathfrak{g}\ltimes\mathfrak{h}$ on itself (and hence the Lie bracket):
$$
[(\xi_1,\eta_1),(\xi_2,\eta_2)] : =\operatorname{ad}_{(\xi_1,\eta_1)}(\xi_2,\eta_2) = ([\xi_1,\xi_2],[\eta_1,\eta_2]+\xi_1\cdot\eta_2 - \xi_2\cdot\eta_1),
$$
where
$$
\xi\cdot\eta := (\dot{\phi}')_\xi\eta = (\dot{\phi}')^\eta\xi = T_{(e_G,e_H)}\phi(\xi,\eta),
$$
(in the final equality thinking of $\phi:G\times H\to H$).
I know this question already have an accepted answer but I want to post my answer here which maybe have a slightly different approach to the question.
We supposed to find an isomorphism $\phi : \text{Lie}(G) \oplus \text{Lie}(H) \to \text{Lie}(G \times H)$. Our fist guess would be the map
$$
\widetilde{\phi} : \mathfrak{X}(G) \oplus \mathfrak{X}(H) \to \mathfrak{X}(G \times H)
$$
defined by $\widetilde{\phi}(X,Y) = X\oplus Y$. This map is linear and preserve Lie bracket , with Lie bracket on $\mathfrak{X}(G) \oplus \mathfrak{X}(H)$ defined as in $(a)$ : for any $(X,Y) ,(X',Y') \in \mathfrak{X}(G) \oplus \mathfrak{X}(H)$ we have
\begin{align*}
\widetilde{\phi}\, \big[(X,Y),(X',Y') \big] &= \widetilde{\phi}\big( [X,X'],[Y,Y'] \big) \\ &= [X,X'] \oplus [Y,Y'] \\ &= [X \oplus Y, X' \oplus Y'] \\ &= [\widetilde{\phi}(X,Y), \widetilde{\phi}(X',Y')].
\end{align*}
So $\widetilde{\phi}$ is a Lie algebra homomorphism . Now we only need to show that the restriction map $ \phi : \text{Lie}(G) \oplus \text{Lie}(H) \to \text{Lie}(G \times H)$ is defined and invertible. If this map is defined (i.e., the image is indeed contained in $\text{Lie}(G \times H)$), then $\phi$ is a Lie algebra isomorphism since $\widetilde{\phi}$ one-to-one and the domain and codomain have the same dimension.
Before show that $\phi$ is defined, I will be a bit pedantic here and remind how vector field $X \oplus Y : G \times H \to T(G \times H)$ defined. For any $(g,h) \in G \times H$ the value $(X \oplus Y)_{(g,h)} \in T_{(g,h)}(G \times H)$ defined as $(X \oplus Y)_{(g,h)} = \alpha^{-1}(X_g,Y_h)$, where
$$
\alpha : T_{(g,h)}(G \times H) \to T_gG \oplus T_hH
$$
is the isomorphism $\alpha(v) := \Big(d(\pi_G)_g(v), d(\pi_H)_h(v)\Big)$.
So now we want to show that $\phi$ is defined, that is for any $X \in \text{Lie}(G)$ dan $Y \in \text{Lie}(H)$, $X \oplus Y$ is a left-invariant vector field. Denote $L_{(g,h)} : G \times H \to G \times H$ as the left translation on the product
$$
L_{(g,h)} (g',h') = (gg',hh') = (L_g\times L_h) (g',h').
$$
Then we must show that for any $(g,h),(g',h')\in G \times H$ we have
$$
d(L_{(g,h)})_{(g',h')}(X \oplus Y)_{(g',h')} =d(L_g \times L_h)_{(g',h')}(X \oplus Y)_{(g',h')} = (X \oplus Y)_{(gg',hh')}.
$$
To show this, as usual denote $\alpha : T_{(g',h')}(G \times H) \to T_{g'}G \oplus T_{h'}H$ as isomorphism $\alpha(v) = \Big(d(\pi_G)_{g'}(v), d(\pi_H)_{h'}(v)\Big)$ and
$\beta : T_{(gg',hh')}(G \times H) \to T_{gg'}G \oplus T_{hh'}H$ as isomorphism $\beta(v) = \Big(d(\pi_G)_{gg'}(v), d(\pi_H)_{hh'}(v)\Big)$. The whole point introducing $\alpha$ and $\beta$ is because the product vector field $X\oplus Y$ defined in terms of this and also for the reason to compute the differential of left-translation on product manifold $L_{(g,h)} = L_g \times L_h$ as we see in the computation below :
\begin{align*}
d(L_g\times L_h)_{(g',h')} (X \oplus Y)_{(g',h')} &=
d(L_g\times L_h)_{(g',h')} \circ \alpha^{-1}(X_{g'},Y_{h'})\\
&=\beta^{-1} \circ \color{blue}{\Big( \beta \circ d(L_g\times L_h)_{(g',h')} \circ \alpha^{-1} \Big)} (X_{g'},Y_{h'}) \\
&=\beta^{-1} \circ \color{blue}{\Big(d(L_g)_{g'}, d(L_h)_{h'} \Big)} (X_{g'},Y_{h'}) \\
&=\beta^{-1}\Big(d(L_g)_{g'}(X_{g'}), d(L_h)_{h'}(Y_{h'}) \Big) \\
&= \beta^{-1}\big( X_{gg'}, Y_{hh'} \big) \\
&= (X \oplus Y)_{(gg',hh')}.
\end{align*}
Therefore $X \oplus Y$ is a left-invariant vector field on $G \times H$ and $\phi : \text{Lie}(G) \oplus \text{Lie}(H) \to \text{Lie}(G \times H)$ is defined. So $\phi (X,Y) = X \oplus Y$ is a Lie algebra isomorphism.
As you can see, without identification, this calculation is so pedantic (which is kind of a bad thing). But this is the only way i know.
Best Answer
The answer to your first question is yes. Let $p: G\to G/H$ denote the projection. Then $\mathfrak{h}$ is a subspace of the kernel of $p_*: \mathfrak g \to Lie(G/H)$. This is because if $X\in \mathfrak h$ then $p_* X = \frac{d}{dt}\vert_{t=0} p(\exp(tX)) = 0$ since $\exp(tX) \in H$ so $p(\exp(tX))$ is constant. By dimensionality reasons, $\mathfrak{h} = \ker p_*$. Thus $p_*$ gives a canonical isomorphism $\mathfrak g/\mathfrak h \to Lie(G/H)$.
I'm really not sure about your second question.