If $F:M\to N$ is a smooth embedding, then so is $dF:TM\to TN$.

Once you know that the image is an immersed submanifold, to show that $F$ is an embedding, it suffices to show that its image is an embedded submanifold; and for that it suffices to show it's embedded in a neighborhood of each point of the image. Let $p\in M$ and $q=F(p)$. Because $F$ is a smooth embedding, it is possible to choose smooth coordinate charts $(U,(x^i))$ containing $p$ and $(V,(y^i))$ containing $q$ such that $F(M)\cap V = F(U)$ and $F|_U$ has a coordinate representation of the form $F(x^1,\dots,x^m) = (x^1,\dots,x^m,0,\dots,0)$. If we let $(x^i, v_i)$ be the associated standard coordinates for $TM$ and $(y^i,w_i)$ those for $TN$, then $dF|_{TU}\colon TU\to TV$ has the coordinate representation $$ dF(x^1,\dots,x^m,v_1,\dots,v_m) = (x^1,\dots,x^m,0,\dots,0,v_1,\dots,v_m,0,\dots,0). $$ The fact that $F(M)\cap V = F(U)$ guarantees that $dF(TM) \cap TV = dF(TU)$, and the coordinate representation above shows that $dF(TU)$ is an embedded $2m$-dimensional submanifold of $TV$.