Does the Riemannian metric induced by a diffeomorphism $F$ exist for a reason other than the existence of vector field pushforwards? My book is Connections, Curvature, and Characteristic Classes by Loring W. Tu (I'll call this Volume 3), a sequel to both Differential Forms in Algebraic Topology by Loring W. Tu and Raoul Bott (Volume 2) and An Introduction to Manifolds by Loring W. Tu (Volume 1).
Definition 1.5 gives the definition for Riemannian metric and Riemannian manifold. Example 1.9 says

If $F : N \to M$ is a diffeomorphism and $< , >$ is a Riemannian metric on
  $M$, then (1.3) defines an induced Riemannian metric $< , >'$ on $N$.



*

*Here $N$ and $M$ are smooth manifolds that hopefully have dimensions.

*Note that the $F_*$ here indeed refers to the differential $F_*,p: T_pN \to T_{F(p)}M$ defined in Volume 1 Section 8.2 and not the latter half $F_*: TN \to TM$ of the bundle map $(F, F_*)$, where $F_*$ is what would be known as $\tilde{F}$ in Volume 1 Section 12.3.
The following is my proof of Example 1.9.


*

*Question 1: Is this proof correct?

*Question 2:


*

*If this proof is correct, then is there a way to do this without relying on pushforwards from Volume 1 or without injectivity of $F$?


*

*I guess we can come up with a similar proof for an embedding, but embeddings are injective. So we'll have to go with investigating local diffeomorphisms, local diffeomorphisms onto image, immersions, etc.

*I'm asking because the Example 1.10 seems to do similarly to Example 1.9 though the $F$ in Example 1.10 is not injective.


*If this proof is incorrect, then why?
Proof:
Notation from Volume 1 Section 2.4: For a smooth manifold $N$, let $\mathfrak X (N)$ be the set of smooth vector fields on $N$, and let $C^{\infty}N$ be the set of smooth functions on $N$ (not germs).
We must show that
A. (Not interested in proving this part, but I'm stating what is to be proven for completeness) For all $p \in N$, the mapping $\langle , \rangle'_p: (T_pN)^2 \to \mathbb R$ is an inner product on $T_pN$, where $\langle , \rangle'_p$ is given as follows:


*

*Let $u,v \in T_pN$. Then $F_{*,p}u, F_{*,p}v \in T_{F(p)}M$.

*Let $\langle , \rangle_{F(p)}: (T_{F(p)}M)^2 \to \mathbb R$ be the inner product on $T_{F(p)}M$ given by the Riemannian metric $\langle , \rangle$ on $M$, at the point $F(p) \in M$.

*Then $(\langle , \rangle'_p)(u,v) = \langle u, v \rangle'_p = \langle F_{*,p}u, F_{*,p}v \rangle_{F(p)}$.
B. $\langle X,Y\rangle' \in C^{\infty}N$ for all $X,Y \in \mathfrak X (N)$, where $\langle X,Y\rangle': N \to \mathbb R$, $\langle X,Y \rangle'(p)=\langle X_p,Y_p\rangle'_p$ $=\langle F_{*,p}X_p,F_{*,p}Y_p\rangle_{F(p)}$.
To prove B:


*

*Let $X,Y \in \mathfrak X (N)$. Then, by Volume 1 Example 14.15, $F_{*}X$ and $F_{*}Y$ are defined vector fields on $M$.

*Hopefully, $F_{*}X$ and $F_{*}Y$ are smooth, i.e. $F_{*}X,F_{*}Y \in \mathfrak X (M)$. (I ask about this step here.)

*$\langle A, B \rangle \in C^{\infty} M$ for all $A,B \in \mathfrak X(M)$, by definition of $\langle , \rangle$ for $M$ (Definition 1.5). 

*$\langle F_{*}X,F_{*}Y \rangle \in C^{\infty}M$, from (2) and (3).

*$\langle X,Y\rangle' = \langle F_{*}X,F_{*}Y \rangle \circ F$, i.e. $\langle X,Y\rangle'$ is the pullback by $F$ of $\langle F_{*}X,F_{*}Y \rangle$

*$\langle X,Y\rangle' \in C^{\infty}N$, by Volume 1 Proposition 6.9, by (4) and by smoothness of $F$.
 A: $\textbf{Question 1:}$ Yes, it is correct.
$\textbf{Question 2:}$ Yes, there is. Even though your proof is correct, it relies more on global properties than it needs to. The trick here is to do things locally, using coordinates.
Let $F\colon M\to N$ be a smooth map and $\left<\cdot\,,\cdot\right>$ be a metric on $N$. You can always define $\left<\cdot\,,\cdot\right>'$ on $M$ the way you did. Then $\left<\cdot\,,\cdot\right>'$ is easily seen to be bilinear and symmetric at each point (please tell me if this is not clear) and, in fact, we can show that it is also smooth (i.e., $\left<X,Y\right>'\colon N\to \mathbb{R}$ is smooth for any $X,Y\in\mathfrak{X}(N)$) without any further assumptions on $F$. After that, all that's left for it to be a metric is to be non-degenerate at each point, which you get by assuming that $(F_*)_p$ is injective at each point $p\in M$ (i.e., assuming $F$ is an immersion), as was already pointed out in the comments.
So let $U\subset M$ be a coordinate neighborhood in $M$ and $V\subset N$ a coordinate neighborhood in $N$ containing $F(U)$, with $\phi=(x^1,\ldots, x^m): U\to\mathbb{R}^m$ and $\psi=(y^1,\ldots, y^n):U\to\mathbb{R}^n$ the corresponding charts. Then for any vector field $\tilde{X}\in\mathfrak{X}(N)$, we have, for $q\in V$
$$\tilde{X}_q=\sum_{i=1}^n\tilde{X}^i\left(q\right)\left(\frac{\partial}{\partial y^i}\right)_q$$
for smooth functions $\tilde{X}^i:V\to\mathbb{R}$.
Furthermore, since the $\frac{\partial}{\partial y^i}$'s form a basis for the tangent space at each point and $\left<\cdot\,,\cdot\right>$ is bilinear, you have functions $g_{ij}:U\to\mathbb{R}$,with $1\leq i,j\leq n$, such that, for any $\tilde{X},\tilde{Y}\in\mathfrak{X}(N)$ and $q\in V$
$$\left<\tilde{X},\tilde{Y}\right>(q)=\sum_{i,j=1}^ng_{ij}(q)\tilde{X}^i(q)\tilde{Y}^j(q)$$
By assumption, this is smooth for every pair of vector fields, so the $g_{ij}$'s must be smooth.
Also, I'm not going to show this, as it's a basic fact of differential geometry (and an expected one too since $F_*$ is supposed to be a generalized derivative), but, for any vector field $X\in\mathfrak{X}(M)$ with
$$X_p=\sum_{i=1}^mX^i(p)\left(\frac{\partial}{\partial x^i}\right)_p$$
you have
$$(F_*)_p(X_p)=\sum_{i=1}^m\sum_{j=1}^nX^i(p)\frac{\partial \tilde{F}^j}{\partial x^i}(p)\left(\frac{\partial}{\partial y^j}\right)_{f(p)}$$
where $\tilde{F}^j=y^j\circ F\circ \phi^{-1}:U\to \mathbb{R}$ for each $1\leq j\leq n$. Then, if $Y\in\mathfrak{X}(M)$ with
$$Y_p=\sum_{i=1}^mY^i(p)\left(\frac{\partial}{\partial x^i}\right)_p$$
you have
$$\left<X,Y\right>'(p)=\sum_{i,j=1}^n\sum_{k,l=1}^mg_{ij}(f(p))X^k(p)\frac{\partial \tilde{F}^i}{\partial x^k}(p)Y^l(p)\frac{\partial \tilde{F}^j}{\partial x^l}(p)$$
which is smooth in $p$ since it's just a sum of products of smooth functions. Since the coordinate neighborhoods are arbitrary, we conclude that $\left<\cdot\,,\cdot\right>'$ is smooth.
More generally, a multilinear map $\omega_q:\left(T_qN\right)^k\to\mathbb{R}$, for each $q\in N$, that varies smoothly with $q$, in the sense that $\omega(X_1,\ldots,X_k):N\to\mathbb{R}$ is smooth for any $X_1,\ldots,X_k\in\mathfrak{X}(N)$, is called a $k$-covariant tensor field and you can show, similarly to what I did above, that $\omega'_p:\left(T_pM\right)^k\to\mathbb{R}$ given by
$$\omega'_p(v_1,\ldots,v_k)=\omega_{f(p)}\left(\left(F_*\right)_pv_1,\ldots,\left(F_*\right)_pv_k\right)$$
varies smoothly with $p$. $\omega'$ is called the pullback of $\omega$ and is usually written $F^*\omega$. What this shows is that, unlike the pushforward, the pullback is always smooth and well-defined without any further assumptions on $F$, other than being smooth.
A: Yes, I think you are basically correct.
Because $F$ is a diffeomorphism, $F:N\to M$ it induces an isomorphism of tangent spaces $F_{*,p}:T_pN\to T_{F(p)}M$. This allows us to define (as you did) an inner product pointwise on $T_pN$ by $\langle u,v\rangle_p=\langle F_* u, F_*v\rangle_{F(p)}$ for any $u,v\in T_pN$. We just need to check that these definitions of inner products $\langle \:\cdot,\cdot\:\rangle$ vary smoothly with $p$ in the sense necessary to define a Riemannian metric. 
To do this, let $X,Y\in \mathfrak{X}(N)$ be given, and notice that $F$ pushes forward smooth vector fields to smooth vector fields (being a $\mathscr{C}^\infty$ diffeomorphism). So, $F_* X ,F_* Y\in \mathfrak{X}(M)$. Then on $N$, 
$$ \langle X,Y\rangle:N\to \mathbb{R}$$
given by $p\mapsto \langle F_*X_p,F_*Y_p\rangle_{F(p)}$ is smooth, being a composition of $p\mapsto F(p)\mapsto \langle F_* X_p, F_* Y_p\rangle_{F(p)}.$ The second map is smooth by one of the characterizations of smoothness of a Riemannian metric, and smoothness of the pushforward vector fields.
