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The pullback bundle is indeed an embedded submanifold of the product $M \times E$. The essential notion here is transversality. (Together with the fact that bundle projection is a submersion).
The full story with all the details is a bit long; I haven't seen it done in any textbook, I found all the steps here and there, and built my own picture of the things.
Definitions
$(1)$ Let $\M,\N$ be a smooth manifolds. Suppose $F:\N \to \M$ is a smooth map, $S \subseteq \M$ is an embedded submanifold. We say $F$ is transverse to $S$ if $\forall x \in F^{-1}\brk{S} \, , \, T_{F\brk{x}}\M= T_{F(x)}S + dF_x(T_x\N)$.
$(2)$ Let $\M,\N,\N'$ be smooth manifolds. Suppose $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. We say $F,F'$ are transverse to each other if $\forall x \in \N, x' \in \N' $ such that $F(x)=F'(x')$ , $T_{F(x)}\M=dF_x(T_x\N) + dF'_{x'}(T_{x'}\N')$.
Note: If either one of $F,F'$ is a submersion, then they are automatically transverse.
Note: Some of ther proofs are at the end of the answer (So it will be possible to skim through the general scheme without all the details at first)
Lemma (1):
$\M,\N$ be a smooth manifolds, $S \subseteq M$ is an embedded submanifold. Let $F:\N \to \M$ be transverse to $S$. Then $F^{-1}(S)$ is an embedded submanifold of $\N$ whose codimension is equal to the codimension of $S$ in $\M$.
proof:
See Theorem 6.30 , in Lee. (pg 144).
Lemma (2): (This is exercise 13 in chapter 6, Lee)
Let $\M,\N,\N'$ be smooth manifolds. Suppose $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. Then $F,F'$ are transverse to each other if and only if the map $F \times F' : \N \times \N' \to \M \times \M$ is transverse to the diagonal $\Delta_\M = \{(x,x)|x \in \M \}$
Lemma (3):
Let $\M$ be a smooth manifold. Then $\Delta_\M = \{(x,x)|x \in \M \}$ is an embedded (smooth) submanifold of $\M \times \M$.
Lemma (4):
Let $\M$ be a manifold. Let $\Delta_\M$ be the diagonal manifold of $\M$. (see Lemma 3 ). Then $T_{\brk{x,x}}\Delta_\M = \text{diag}\brk{T_x\M \times T_x\M}=\{(v,v)| v \in T_xM \}$. (i.e the tangent space of the diagonal is the diagonal of the tangent space).
proof:
Since any tangent vector can be realized as a derivative of a path, the tangent space to a manifold is identical to the set of derivatives of paths. Since $\Delta_\M$ is an embedded submanifold, a path $\be:I \to \Delta_\M $ is smooth if and only if it is smooth when considered as a path into the product $\M \times \M$ if and only if each of its components is smooth. So $\be(t)=\brk{\al\brk{t},\al\brk{t}}$ , where $\al : I \to \M$, so $\dot \be (0) \overset{(*)}= \brk{\dot \al (0),\dot \al (0)}$ , hence its clear the tangent space to the diagonal is exactly the diagonal of the tangent space. (Where in (*) we used the canonical isomorphism between $T_{(x,x')}\brk{\M \times \M'} = T_x\M \oplus T_{x'}\M'$ via the differentials of the projections onto the different components).
corollary (1):
Let $\M,\N,\N'$ be smooth manifolds, $F:\N \to \M \, , \, F':\N' \to \M$ are smooth maps. (In short we write $\N\overset{F}{\rightarrow}\M\overset{F'}{\leftarrow}\N'$). Assume $F,F'$ are transverse to each other. Then the fiber product of this diagram, which is defined as $\{\brk{x,x'} \in \brk{\N, \N'}|F(x)=F'(x') \}$ is an embedded smooth submanifold of the product $\N \times \N'$.
proof of corollary (1):
The fibered product $\N \times_{\M} \N'$ is the inverse image $(F \times F')^{-1}\brk{\Delta_{\M}}$. By Lemma 3, $\Delta_\M$ is a submanifold of $\M \times \M$. Now combine Lemma 2 and Lemma 1.
corollary (2):
Let $\M,\N,\N'$ be smooth manifolds, $\N\overset{F}{\rightarrow}\M\overset{F'}{\leftarrow}\N'$. If either one of $F,F'$ is a submersion, then the fiber product $\N \times_\M \N' = \{\brk{x,x'} \in \brk{\N, \N}|F(x)=F'(x') \}$ is an embedded submanifold of the product $\N \times \N'$.
proof of corollary (2):
If one of $F,F'$ is a submersion, then these two maps are automatically transverse to each other. Now use corollary (1).
In particular we get the following proposition:
Let $\pi: E \to B$ be a vector bundle, $f:B' \to B$. The pullack bundle $f^*\brk{E}$ is an embedded submanifold of the product $B' \times E$.
(This is becuse the bundle projection $\pi$ is always a submersion).
proof of Lemma (2):
First, we need a sublemma:
Sub-lemma:
Let $V$ be a vector space, $V_1,V_2 \subseteq V$ are subspaces. Let $\text{diag}(V \times V) = \{(v,v)|v \in V \} $. Then
$V \oplus V = \text{diag}(V \times V) + \brk{V_1 \oplus V_2} \iff V = V_1 + V_2$
proof of the sublemma:
$\Rightarrow :$ Let $v \in V$. Then $(v,0) \in V \oplus V$, hence by our assumption $\exists \til v \in V, v_1 \in V_1 , v_2 \in V_2$ such that $(v,0) = (\til v ,\til v) + (v_1,v_2)=(\til v + v_1, \til v + v_2) \Rightarrow \til v = -v_2, v = \til v + v_1 = v_1 - v_2 \in V_1 + V_2 $ .
$\Leftarrow :$ Note that both sides of the left equation are subspaces. Hence, from symmetry it's enough to show that $\forall v \in V \, , \, (v,0) \in \text{diag}(V \times V) + \brk{V_1 \oplus V_2}$. The assumption $V =V_1 + V_2 \Rightarrow \exists v_i \in V_i$ such that $v = v_1 -v_2$. Define $\til v = -v_2$, so we get $(v,0)=(v_1-v_2,\til v +v_2)=(v_1 + \til v, v_2 + \til v) = (\til v,\til v) +(v_1,v_2)$.
Now to the actual proof of Lemma (2):
By definition (1), $F \times F'$ is transverse to the diagonal if
\begin{split}
&\forall (x,x') \in (F \times F')^{-1}\brk{\Delta_\M} \, , \, T_{(F \times F')\brk{x,x'}}\brk{\M \times \M}= T_{(F \times F')(x,x')}\Delta_\M + d(F \times F')_{(x,x')}(T_{(x,x')}\brk{\N \times \N'}) \iff \\
& T_{\brk{F(x),F'(x')}}\brk{\M \times \M}= T_{\brk{(F(x),F'(x')}}\Delta_\M + d(F \times F')_{(x,x')}(T_x\N \oplus T_x\N') \iff \\
& T_{\brk{F(x),F(x)}}\brk{\M \times \M}= T_{\brk{(F(x),F(x)}}\Delta_\M + d(F \times F')_{(x,x')}(T_x\N \oplus T_x\N') \iff \\
&T_{\brk{F\brk{x}}}\M \oplus T_{\brk{F\brk{x}}}\M \overset{Lemma 4}= \text{diag}\brk{T_{F(x)} \M \times T_{F(x)}\M} + \brk{ dF_x \brk{T_x \N} \oplus dF'_{x'} \brk{T_{x'}\N'}} \overset{Sub-lemma} \iff \\
&T_{F(x)}\M =dF_x \brk{T_x \N} + dF'_{x'} \brk{T_{x'}\N'}
\end{split}
Since the last row is the defintion transverse maps, we finished.
proof of Lemma (3):
The diagonal is the graph of the smooth function $Id_\M$, and graphs of smooth functions are always embedded submanifolds of the product of the domain and the codomain. (See prop 5.4, Lee).
Best Answer
As you wrote, let $f^*E := \{(e,n) \in E \times N| \ \pi(e)=f(n) \}$. I like to to it the other way around, so let our trivialisations for $E$ be given by $$\phi: \pi^{-1}(U) \to U \times \mathbb{R}^k, \ e \mapsto (\pi(e), pr_2(\phi(e)))$$ where $pr_2: U \times \mathbb{R}^k \to \mathbb{R}^k, (x,v) \mapsto v$ (why can we write it that way?). Now, let (as you wrote) $$\rho: f^*E \to N, (e,n) \mapsto n$$ wherefore: $$\rho^{-1}(V) = \{(e,n) \in E \times V| \ \pi(e) = f(n) \}.$$ Thus we can define the trivialisation (why?): $$\psi: \rho^{-1}(f^{-1}(U)) \to f^{-1}(U) \times \mathbb{R}^k, (e,n) \mapsto (n, pr_2(\phi(e)))$$ again, $pr_2$ ist just the projection onto $\mathbb{R}^k.$