Let $H$ be a normal subgroup of a group $G$ and $p$ the canonical projection of $G$ onto $G/H$. Let $t$ be a section of $p$. The map $\phi:G\rightarrow H$ defined by $\phi(g)=gt(p(g^{-1}))$ is well defined.
Is $\phi$ a group homomorphism?
I don't think that but I don't know why. Could anyone please help?
Thank you in advance.
What you have makes no sense: $p$ has domain $G$ and codomain $G/H$; so you cannot calculate $p\circ p$: the output of $p$ is a coset of $H$, and the input of $p$ are elements of $G$, so you cannot do the composition.
To define something like this you need a function $f\colon G/H\to G$ so that you can instead look at $$ x f\bigl( p(x^{-1})\bigr).$$ In order for this to lie in $H$, we need $$eH = p(x)p(f(p(x^{-1}))),$$ and this can be achieved by requiring that $p\circ f = \mathrm{id}_{G/H}$.
A function $f\colon G/H\to G$ such that $p\circ f=\mathrm{id}_{G/H}$ is called a section of $p$; the image of $f$ is a complete set of coset representatives of $H$ in $G$. It is true that for all $x\in G$, $xf(p(x^{-1}))$ lies in $H$; but in general there are many sections (as many as complete sets of coset representatives). Note in particular that a section is not required to send $e$ to $e$; just to some element of $H$.
The section $f$ can be chosen to be a group homomorphism if and only if $G$ is a semidirect product $G=H\rtimes Q$ for some subgroup $Q$; namely, if $G=H\rtimes Q$, then for each $g\in G$ there exists a unique $q\in Q$ such that $gH=qH$, and we can define $f\colon G/H\to G$ by $f(gH)=q$; and conversely if $f$ is a homomorphism, then we can let $Q=\mathrm{Im}(f)$.
Let $f$ be a fixed section of $p\colon G\to G/H$; and let $\phi_f\colon G\to H$ be the function given by $\phi_f(x) = xf(p(x^{-1}))$. If $f(e)\neq e$ then $\phi_f$ is definitely not a homomorphism, since $\phi_f(e) = f(e)$. So we can require the section to send $e_{G/H}$ to $e_G$. Then $\phi_f(h) = h$ for all $h\in H$.
But even then, and even if $f$ is a homomorphism, the resulting map need not be a homomorphism. Here is an easy example: take $G=S_3$, $H=A_3$. To get a homomorphism $\phi\colon S_3 \to A_3$, we need $\phi$ to send all elements of order $2$ to the trivial element of $A_3$ (since $A_3$ is cyclic of order $3$). But that means the kernel contains all transpositions, and therefore is the trivial morphism; that is, the only homomorphism $\phi\colon S_3\to A_3$ is the trivial map. But $\phi_f$ is not trivial for any $f$ that sends $e_{G/H}$ to $e_G$, as noted above: it restricts to the identity on $A_3$. Thus, no choice of $f$ will make $\phi_f$ a homomorphism.
Added: In response to the comment claiming the map will be a homomorphism when $G$ is abelian, the claim is also incorrect.
Let $G=\mathbb{Z}/4\mathbb{Z}$, $H=\{\overline{0},\overline{2}\}$; then $G/H\cong C_2$ the cyclic group of order $2$. Define $f\colon C_2\to G$ to be the section given by $f(e)=\overline{0}$ (required, as above), and $f(x)=\overline{1}$ (a similar argument holds if $f(x)=\overline{3}$). Calculating, we have: $$\begin{align*} \phi_f(\overline{0}) &= \overline{0}+f(p(\overline{0})) = \overline{0}+f(e) = \overline{0}+\overline{0}=\overline{0}\\ \phi_f(\overline{1}) &= \overline{1}+f(p(\overline{3})) = \overline{1}+f(x) = \overline{2}\\ \phi_f(\overline{2}) &= \overline{2}+f(p(\overline{2})) = \overline{2}\\ \phi_f(\overline{3}) &= \overline{3}+f(p(\overline{1})) = \overline{3}+\overline{1}=\overline{0}. \end{align*}$$ In particular, the elements mapping to $\overline{0}$ do not form a subgroup, and $\phi_f(\overline{1}+\overline{1}) = \overline{2} \neq \phi_f(\overline{1})+\phi_f(\overline{1})$.