$\textbf{Question 1:}$ 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.
Best Answer
As long as $\varphi \colon M \to N$ is differentiable, the map $d\varphi_p$ is a vector space homomorphism from $T_pM$ to $T_{\varphi(p)}N$ for any $p \in M$. So $\phi$ pushes tangent vectors forward without needing invertibility.
But if a vector field $X$ on $M$ is to push forward to a vector field $\varphi_* X$ on $N$, we need a vector $(\varphi_* X)_q$ in every $T_qN$. There's no way to do this unless $q$ is in the image of $\varphi$. So we require $\varphi$ to be surjective. Then if $\phi(q) = q$, we can define $(\varphi_* X)_q = d\varphi(X)_p$ In order for this to be well-defined, we need for $\varphi$ to be one-to-one. So we can say $$ (\varphi_* X)_q = d\varphi(X)_{\varphi^{-1}(q)} $$ I assume you need $\varphi^{-1}$ to be differentiable in order for $\varphi_*X$ to be smooth.
As examples, consider $M = \mathbb{R}^2$ with coordinates $(x_1,x_2)$, $N = \mathbb{R}$ with coordinate $y$, and $\varphi(x_1,x_2) = x_1$. Then clearly $d\varphi$ takes $\frac{\partial}{\partial x_1}$ to $\frac{\partial}{\partial y}$ and $\frac{\partial}{\partial x_2}$ to $0$. What should $\varphi_*\left(x_2 \frac{\partial}{\partial x_1}\right)$ be though? You'd have to choose a lifting $\psi\colon N \to M$ (in other words, a function $x_2(y)$), which seems problematic from a naturality standpoint.
Or $M = N = \mathbb{R}$ with $\varphi(x) = x^3$. Let $X = \frac{\partial}{\partial x}$. Then $d\varphi(x,\frac{\partial}{\partial x}) = (x^3,3x^2\frac{\partial}{\partial y})$ To define $\varphi_*x X$ at a point $y \in N$, you'd need to take $d\varphi(y^{1/3},\frac{\partial}{\partial x}) = (y,3y^{2/3}\frac{\partial}{\partial y})$. This isn't going to be smooth at $y=0$.
You can read more about this on Wikipedia.