Surely you mean "former to latter"?
I think the natural map is injective. Let $V$ and $W$ have
bases $v_1,\ldots,v_r$ and $w_1,\ldots,w_s$ respectively.
Let $g_1,\ldots,g_t$ be coset representatives for $H$ in $G$.
Then a basis for $\mathrm{Ind}_H^G V\otimes \mathrm{Ind}_H^G W$
consists of the $(v_i g_k)\otimes(w_j g_l)$. The image of
the natural injection from $\mathrm{Ind}_H^G(V\otimes W)$
is spanned by the $(v_i g_k)\otimes(w_j g_l)$ with $k=l$.
There are exactly the right number of these.
Not always! However, the closure is a polyhedron.
Not always: Take $P = \{a | 0 \leq a \leq 1\}$. Take $Q= \{ b,c | b\geq 0, c=1\}$. Then under the map $x=ab$, $y=ac$. Since $P$ is the convex hull of $(0)$ and $(1)$, and $Q$ is the ray starting at $(0,1)$ and going in direction $(1,0)$, $P \otimes Q$ is the convex hull of $(0,0)$ and the ray starting at $(0,1)$ and going in direction $(1,0)$, which is:
$P \otimes Q = \{x,y | 0\leq x, 0 \leq y \leq 1, (y>0 \vee x=0) \}$
This not a polyhedron because it is not closed.
The closure is: Define $B$ to be the following convex body. First we show that $B$ is a polyhedron. Next we will show that $cl(P \otimes Q)=B$.
$B= \operatorname{conv}(P_p \otimes Q_p + ((P_p+P_c) \otimes Q_c) + (P_c \otimes (Q_p+Q_c)) + (P \otimes Q_l) + (P_l \otimes Q)) $
Since $\operatorname{conv}(A + B) = \operatorname{conv}(A) + \operatorname{conv}(B)$,
$B= \operatorname{conv}(P_p \otimes Q_p) + \operatorname{conv}(((P_p+P_c) \otimes Q_c) + (P_c \otimes (Q_p+Q_c)) )+ \operatorname{conv}((P \otimes Q_l) + (P_l \otimes Q))) $
The first part is clearly a polytope. The last part is clearly a linear subspace. The cone is the tricky but one can view a cone as the convex hull of finitely many rays. A ray tensor a polytope is the convex hull of finitely many rays, thus a cone. A ray tensor a cone is the convex hull of finitely many rays, thus a cone again. Taking the convex hull of different cones could produce more linear subspaces but will not take you out of the world of polyhedra. (Dima might be able to produce a better argument than this?)
Next we show that $cl(P \otimes Q) \subseteq B$. As a polytope, it is convex and closed, so it is enough to show that any element in $P$ tensor an element in $Q$ is in $B$. But this is obvious - just split that element into a sum.
Finally we show that $B \subseteq cl(P \otimes Q)$. Since $P \otimes Q$ is convex, $cl(P\otimes Q)$ is convex, so need to show that a sum of an element in $P_p \otimes Q_p$, an element in $(P_p+P_c)\otimes Q_c$, etc. is in $B$. Assume for simplicity we merely need to add $a \otimes b$ in $P_p \times Q_p$ to $c \otimes d$ in $(P_p+P_c) \otimes Q_c$. (To get the general caes, you just repeat the argument). Let $e$ be any element of $Q_p$. Then we notice that
$\lim _ {\lambda \to 0} \left((1-\lambda) \left[a \otimes b\right] + \lambda \left[c \otimes \left(\frac{d}{\lambda}+ e \right)\right]\right)= a\otimes b+ c\otimes d $
$a \otimes b$ and $c \otimes \frac{d}{\lambda}+ e$ are in $P \otimes Q$ as long as $\lambda>0$, so a convex combination of them is as well, so the limit as $\lambda \to 0$ is in $cl(P \otimes Q)$. The key fact is that $Q_c$ is closed under multiplication by positive real numbers. Since $P_c$, $Q_l$, and $P_l$ are as well, we can apply this trick again to get an arbitrary sum.
Best Answer
Suppose $\sum f_i\otimes g_i$ is in the kernel and assume that the $g_i$ are linearly independent. For every $a\in A$ and $\lambda\in C^*$, we have $\sum\lambda(f_i(a))g_i=0$, so by assumption $\lambda(f_i(a))=0$ for all $i$. Since $a$ and $\lambda$ were arbitrary, this implies $f_i=0$ for all $i$.