$\newcommand{\Hom}{\operatorname{Hom}}$Here is an elaboration of my comment with what happens in characteristic $2$ when $d=3$:
The usual decomposition rule gives us a filtration of $S^dV\otimes V$ with two factors: $H^0(4)$ and $H^0(2)$ (I use the notation from Jantzen's Representations of Algebraic Groups and write all weights in terms of the fundamental weight).
Applying the Jantzen sum formula, we see that $H^0(4)$ has a composition series consisting of $L(4)$, $L(2)$ and $L(0)$.
We also see that $H^0(2)$ has a composition series consisting of $L(2)$ and $L(0)$.
All this gives us a composition series $$0 \subseteq M_1 \subseteq M_2 \subseteq M_3 \subseteq M_4 \subseteq M_5 = S^3V\otimes V$$ where $M_1\cong L(2)$, $M_2/M_1\cong L(0)$, $M_3/M_2\cong L(4)$ and $\{M_4/M_3,M_5/M_4\}\cong \{L(2),L(0)\}$. Which order the two top factors come in is less obvious (I will need to think a bit about it), and whether we actually have $S^3V\otimes V\cong H^0(4)\oplus H^0(2)$ I will also need to think a bit more to figure out.
Added: So, after some further thought, we can actually say a bit more.
First note that as mentioned by Jim Humphreys, we have $S^3V\otimes V\cong L(1)\otimes L(1)\otimes L(1)^{(1)}$ which means that it is self-dual. In particular, we see that our composition series can be chosen to be "symmetric", so we get $M_4/M_3\cong L(0)$ and $M_5/M_4\cong L(2)$ (it is also good to notice that we actually have $M_2\cong H^0(2)$ and $M_5/M_2\cong H^0(4)$ as these are sometimes easier to work with).
We can also show that $S^3V\otimes V$ is indecomposable. In fact, we have $\operatorname{soc}_{SL_2}(S^3V\otimes V) = L(2)$.
To see this, we need a bit more machinery (it might be possible to do this in a more elementary way). Let $G = SL_2$ and let $G_1$ be the first Frobenius kernel of $G$. We let $\lambda = \lambda_0 + p\lambda_1$ be a dominant weight with $\lambda_0 < p$ and use that $L(\lambda) \cong L(\lambda_0)\otimes L(\lambda_1)^{(1)}$. Now we note that $$\Hom_G(L(\lambda),L(1)\otimes L(1)\otimes L(1)^{(1)})$$ $$\cong \Hom_{G/G_1}(L(\lambda_1)^{(1)},\Hom_{G_1}(L(\lambda_0),L(1)\otimes L(1))\otimes L(1)^{(1)})$$ so it is sufficient to show that $\operatorname{soc}_{G_1}(L(1)\otimes L(1)) = L(0)$.
To see this we further note that it will suffice to show that $\operatorname{soc}_G(L(1)\otimes L(1)) = L(0)$ since the $G_1$-socle is a $G$-submodule. But this final part is a simple calculation, as we clearly just need to check that neither $L(1)$ nor $L(2)$ are submodules. That $L(1)$ is not a submodule is clear by parity (all highest weights of composition factors in $L(1)\otimes L(1)$ must be even), and that $L(2)$ is not a submodule is seen by noting that $$\Hom_G(L(2),L(1)\otimes L(1))\cong \Hom_G(L(1),L(1)\otimes L(2))\cong \Hom_G(L(1),L(3))$$ and $L(3)$ is simple (it is the 2'nd Steinberg module as also mentioned by Jim Humphreys).
A few final notes: The above actually shows that as a $G_1$-module, $L(1)\otimes L(1)$ is the injective hull of the trivial module. This is a general fact about $SL_2$ in characteristic $2$, ie, that for all $r$, $St_r\otimes St_r$ is the injective hull of the trivial module as a $G_r$-module (this does not generalize to other groups, nor to other primes).
Also, the conclusion about the module $S^3V\otimes V$ is in fact that it is indecomposable tilting (in the notation from Jantzen, it is denoted $T(4)$).
Best Answer
If $k$ is of characteristic zero or characteristic $p$ and $G$ is a pro-$p'$-group, then yes. In that case, everything in sight is semisimple by Maschke's theorem, the eigenvalues $a_i$ and $b_j$ give you the decomposition into irreducibles, and every pair of equal eigenvalues contributes one degree of freedom by Schur's lemma.
Otherwise no, even in the simplest case, because of lack of semiplicity. If $k=\overline{\mathbb{F}_p}$, $G=C_p$, $\rho_V(\phi) = \begin{pmatrix}1&1\\&1\end{pmatrix}$ and $\rho_W(\phi)=\begin{pmatrix}1&\\&1\end{pmatrix}$, then $a_1=a_2=b_1=b_2$ so that your conjecture would predict $\dim_k \operatorname{Hom}_G(V,W) = 4$, but it is only 2, because $V$ is not semi simple, but $W$ is, and therefore every morphism must factor through $V/\operatorname{rad}(V)$.