$\newcommand{\Hom}{\operatorname{Hom}}$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 $T(4)$).