My differential geometry is a bit rusty, so I'd like some help with what follows:
I have the following setting: $M,N$ Riemannian manifold of dimension $m<n$ with codimension $d$, $M$ in embedded in $N$ via an isometry $\phi$. We define a new manifold $\mathcal{T}=M\times B^d\ni(p,x)$ and a map: $$\begin{array}{rclc}\overline\phi:&\mathcal{T}&\longrightarrow&N\\&(p,x)&\longmapsto&\exp_{\phi(p)}x\end{array}$$ Assume this new map is again an embedding. Take a vector: $$v=(v_M,v_B)\in T_{(p,x)}\mathcal{T}\cong T_pM\times T_xB^d$$ I would like to know what is: $$\overline\phi_*v\in TN$$ We are taking the pushforward of an exponential map, so intuitively I would say that it is just the parallel transport of $\overline\phi_*v$, where $v$ is now seen as an element of $T_{(p,0)}\mathcal{T}$, along the geodesic $\gamma(t)=\exp_{\phi(p)}(tx)$.
Is this correct? If it is, what would be the easiest way to prove it? If not so, is there a nice expression/interpretation for it?
Edit: Some details on the definition of $\overline\phi$. As written above, the map doesn't really make sense, because $\exp_{\phi(p)}$ should take an element of $T_\phi(p)N$ as argument, while $x\in B^d$ (thanks Jack Lee). However, assuming $M$ is orientable, we have a natural identification of $\mathcal{T}$ as a submanifold of $NM$ (the normal bundle to $M$ in $N$; fix a basis of $NM$...), in particular we can see $(p,x)$ as an element of $N_{\phi(p)}M\subset T_{\phi(p)}M$.