Let $G$ and $H$ be lie groups and $\Gamma_f:=\{(g,f(g) |g \in G\} \subset G \times H$. Let furthermore $p_1: G \times H \to, (g,h) \mapsto g$ be the projection and $p:= p_1 |_{\Gamma_f}: \Gamma_f \to G$.
Claim: I wanna show that $p_*:T_{(e,e)}\Gamma_f \to T_eG$ is injective.
Proof: Let $(X,Y) \in T_{(e,e)}\Gamma_f$ with $p_*(X,Y)=0$. Hence, we have that $X=0$. Now let $\alpha(t)$ be the one parameter group $\alpha: \Bbb R \to \Gamma_f$ for $(0,Y)=(X,Y)$ $\Rightarrow$ $(p \circ \alpha)(t)$ is the one parameter group for $0 \in T_eG$ $\Rightarrow$ $(p \circ \alpha)(t)$ is constant $\Rightarrow$ $\alpha(t)$ is constant.
How can we conclude that $(p \circ \alpha)(t)$ is constant and $\alpha(t)$ is constant? Also what does it means that it is the one-parameter group for $(0,Y)$. I don't see it. Can someone please help me?
If $(X, f_*X) \in \ker p_*$, then $X=0$, so $f_*X=0$ as well. You only need to know that $T_{(e,e)}\Gamma_f = \{(X, f_*X)\in T_{(e,e)}(G \times H) \mid X \in T_eG\}$. You don't need to talk about one-parameter subgroups at all here.
This is a general fact: if $f:M \to N$ is a smooth map between manifolds, then $\Gamma_f = \Phi^{-1}[\Delta]$, where $\Delta \subseteq N\times N$ is the diagonal and $\Phi:M\times N \to N \times N$ is given by $\Phi(x,y)=(f(x),y)$. Then the tangent space $T_{(x,f(x))}\Gamma_f$ is the inverse image of the tangent space $T_{(f(x),f(x))}\Delta$ under the derivative $d\Phi_{(x,f(x))}(v,w) = (df_xv, w)$ meaning that if $(v,w)$ is tangent to $\Gamma_f$, then $w$ is the image of $v$ under $df_x$.