Let $M,N$ be smooth manifolds of dimensions $m,n$ respectively and $\phi:M\to N$ be a smooth map. Let $g$ be a metric on $N$. If $\mathbf{x}$ and $\mathbf{y}$ are local coordinates on $M$ and $N$ respectively then the metric takes the form: $$g=g_{ij}dy^i\otimes dy^j,\;\;\;\;\;i,j=1,\ldots,m.$$ How can one define the pullback of $g$ given that I know the pullback of 2-forms (and $k$-forms in general)?
My Try Lets work with 2 forms for simplicity. The pullback of a two form $\alpha$ is defined to be $$F^*(\alpha)(v_1,v_2)=\alpha(F_*v_1,F_*v_2),$$ where $v_1,v_2$ are tangent vectors at some point $p\in M$ and $F_*$ is the pushforward map. Now suppose that $\alpha=\beta\otimes\gamma$ where $\beta,\gamma$ are 1-forms on $N$. Using the above I can define the map $$F^*(\beta\otimes\gamma)(v_1,v_2)=(\beta\otimes\gamma)(F_*v_1,F_*v_2)=\beta(F_*v_1)\cdot\gamma(F_*v_2).$$ But $\beta(F_*v_1)=F^*\beta(v_1)$ and $\gamma(F_*v_2)=F^*\gamma(v_2)$ and $$ (F^*\beta\otimes F^*\gamma)(v_1,v_2)=F^*\beta(v_1)\cdot F^*\gamma(v_2)$$ so that I get $$ F^*(\beta\otimes\gamma)=F^*\beta\otimes F^*\gamma.$$ Which looks like a 2-form on $M$. So given this $$F^*(g)=F^*(g_{ij}dy^i\otimes dy^j)=(g_{ij}\circ F)\;F^*(dy^i)\otimes F^*(dy^j)=(g_{ij}\circ F)\frac{\partial F^i}{\partial x^k}dx^k\otimes\frac{\partial F^j}{\partial x^\ell}dx^\ell,$$ which looks like a metric. Is my reasoning consistent?
Yes, everything you said is sound. One can define pull-backs of arbitrary covariant tensor fields by smooth maps, the tensor being alternating is completely immaterial at this point. Given a 2-covariant tensor $g$ in the target manifold, one can consider the 2-covariant tensor field $F^*g$ in the source manifold. You have a correct coordinate expression for $F^*g$. I would probably just rewrite it as $$F^*g = (g_{ij}\circ F)\frac{\partial F^i}{\partial x^k}\frac{\partial F^j}{\partial x^\ell}\,{\rm d}x^k\otimes {\rm d}x^\ell,$$to emphasize the tensor components.