[proofplan]
The hypotheses defining the optimal control problem are used through the stated assumption that the normal Pontryagin maximum principle applies to this problem. We specialize its endpoint transversality condition to the Hamiltonian convention $H=p\cdot f-L$ and cost multiplier $1$, which gives the endpoint covector $p(t_1)+d\Phi_{x^*(t_1)}$. Since the terminal constraint is $x(t_1)\in M$, admissible first-order endpoint variations lie in the tangent space $T_{x^*(t_1)}M$, so the endpoint covector annihilates that tangent space. Finally, the Euclidean [inner product](/page/Inner%20Product) identifies $d\Phi_{x^*(t_1)}$ with $\nabla\Phi(x^*(t_1))$, converting annihilation of the tangent space into membership in the stated normal space.
[/proofplan]
[step:Extract the endpoint transversality condition from the normal maximum principle]
Define the terminal point $y_1\in M$ by $y_1:=x^*(t_1)$. Let $T_{y_1}M\subset\mathbb{R}^n$ denote the tangent space of $M$ at $y_1$, and let $(\mathbb{R}^n)^*$ denote the dual [vector space](/page/Vector%20Space) of linear maps from $\mathbb{R}^n$ to $\mathbb{R}$. Let $d\Phi_{y_1}:\mathbb{R}^n\to\mathbb{R}$ denote the differential of $\Phi$ at $y_1$. Let $p(t_1)^\flat\in(\mathbb{R}^n)^*$ denote the covector $w\mapsto p(t_1)\cdot w$. Because the normal Pontryagin maximum principle is assumed to apply to the stated optimal control problem with cost multiplier $1$ and with Hamiltonian convention
\begin{align*}
H(t,x,u,p)=p\cdot f(t,x,u)-L(t,x,u),
\end{align*}
its terminal manifold transversality clause says that the endpoint covector
\begin{align*}
p(t_1)^\flat+d\Phi_{y_1}\in (\mathbb{R}^n)^*
\end{align*}
annihilates every feasible first-order terminal variation tangent to $M$ at $y_1$.
For the constraint $x(t_1)\in M$, those first-order terminal variations are precisely vectors in the tangent space $T_{y_1}M$. Hence, for every $w\in T_{y_1}M$,
\begin{align*}
\bigl(p(t_1)^\flat+d\Phi_{y_1}\bigr)(w)=0.
\end{align*}
[guided]
Let $y_1:=x^*(t_1)$ denote the terminal point of the minimizing trajectory. Let $T_{y_1}M\subset\mathbb{R}^n$ denote the tangent space of $M$ at $y_1$, let $(\mathbb{R}^n)^*$ denote the dual vector space of linear maps from $\mathbb{R}^n$ to $\mathbb{R}$, let $d\Phi_{y_1}:\mathbb{R}^n\to\mathbb{R}$ denote the differential of $\Phi$ at $y_1$, and let $p(t_1)^\flat\in(\mathbb{R}^n)^*$ denote the covector $w\mapsto p(t_1)\cdot w$. The endpoint constraint in the problem is not that $x(t_1)$ is fixed; it is that $x(t_1)$ lies on the embedded submanifold $M$. Therefore the allowed first-order endpoint displacements are exactly the tangent vectors $w\in T_{y_1}M$.
The normal Pontryagin maximum principle is being used here as an assumption, not proved from scratch. The regularity of $f$, $L$, and $\Phi$, the fixed initial condition, the terminal manifold constraint, and local minimality are precisely the data placing the minimizer in the scope of that assumed maximum principle. Under the Hamiltonian convention
\begin{align*}
H(t,x,u,p)=p\cdot f(t,x,u)-L(t,x,u),
\end{align*}
and with cost multiplier equal to $1$, the endpoint contribution in the first variation is represented by the covector
\begin{align*}
p(t_1)^\flat+d\Phi_{y_1}\in(\mathbb{R}^n)^*.
\end{align*}
The sign is tied to this Hamiltonian convention: the running cost enters $H$ with a minus sign, so the terminal cost appears with the plus sign shown above.
The transversality clause says that this endpoint covector must vanish on all tangent endpoint variations allowed by the terminal constraint. Since the feasible terminal variations are the vectors in $T_{y_1}M$, we get
\begin{align*}
\bigl(p(t_1)^\flat+d\Phi_{y_1}\bigr)(w)=0
\end{align*}
for every $w\in T_{y_1}M$.
[/guided]
[/step]
[step:Identify the endpoint covector with the Euclidean vector in the statement]
Use the Euclidean inner product on $\mathbb{R}^n$ to identify covectors and vectors. Under this identification, the covector $d\Phi_{y_1}$ corresponds to the gradient vector $\nabla\Phi(y_1)$, and the covector $p(t_1)^\flat$ is represented by the vector $p(t_1)$. Therefore the previous annihilation condition becomes
\begin{align*}
\bigl(p(t_1)+\nabla\Phi(y_1)\bigr)\cdot w=0
\end{align*}
for every $w\in T_{y_1}M$.
[/step]
[step:Convert annihilation of the tangent space into normality]
By definition of the Euclidean normal space to $M$ at $y_1$,
\begin{align*}
N_{y_1}M=\{v\in\mathbb{R}^n: v\cdot w=0\text{ for every }w\in T_{y_1}M\}.
\end{align*}
The vector $p(t_1)+\nabla\Phi(y_1)$ satisfies exactly this defining condition. Hence
\begin{align*}
p(t_1)+\nabla\Phi(y_1)\in N_{y_1}M.
\end{align*}
Substituting back $y_1=x^*(t_1)$ gives
\begin{align*}
p(t_1)+\nabla\Phi(x^*(t_1))\in N_{x^*(t_1)}M,
\end{align*}
which is the asserted terminal transversality condition.
[/step]
