No, $Ric > 0$ does not imply $K >0$ or even $K\geq 0$, except in dimensions at most $3$.
For a counterxample to $Ric > 0\implies K > 0$, one can look at products. For example, $S^2\times S^2$ equipped with the product of round metrics has positive Ricci curvature, but each point has $0$ curvature planes.
Unfortunately, I don't know of an explicit counterexample to $Ric > 0\implies K\geq 0$, but here are two reasons that such examples must exist.
First, note that $Ric > 0$ is an open condition - small perturbations of the metric preserve this property. On the other hand, $K\geq 0$ is a closed condition, so one should expect that small perturbations of $K\geq 0$ ruin this property. So, it should be possible to slightly deform the product metric on $S^2\times S^2$ to get a counterexample. (I admit this is not a proof, just an idea).
Now, here is a proof that such examples must exist. We begin with a theorem of Sha and Yang (see here) that any number of connect sums of $S^3\times S^4$ with itself admit a metric of positive Ricci curvature.
On the other hand, a theorem of Gromov (see here) asserts a Betti number bound (depending only on the dimension) in the presence of non-negative sectional curvature.
Now, if we connect sum enough $S^3\times S^4$s together, we can achieve arbitrarily large Betti numbers. By Gromov's result, such a manifold does not admit a metric with $K\geq 0$, but by Sha and Yang's result, such a manifold does admit a metric with $Ric > 0$.
The fact that $X$ and $Y$ are orthonormal says nothing about whether or not $X_1$ and $X_2$ are orthonormal. Thus, the condition $\langle R(X_1,Y_1)Y_1, X_1\rangle =1$ does not need to hold. In fact, $X_1$ and $Y_1$ could be linearly dependent, in which case the curvature is $0$.
Best Answer
$\DeclareMathOperator{\Ric}{Ric}$For uniformity, let's denote our Riemannian factors as $(M_{i}, g_{i})$ for $i = 1$, $2$, and denote by $\pi_{i}:M_{1} \times M_{2} \to M_{i}$ the projections. The product metric is $$ g = \pi_{1}^{*}g_{1} + \pi_{2}^{*}g_{2}, $$ which we might denote $g_{1} \oplus g_{2}$ or $g_{1} \times g_{2}$ (depending whether we view $TM$ as $\pi_{1}^{*}TM_{1} \oplus \pi_{2}^{*}TM_{2}$ or as $TM_{1} \times TM_{2}$), and similarly for the Ricci tensor.
If $X_{1}$ is a vector field on $M_{1}$, there is a unique vector field $\widehat{X}_{1}$ on $M$ satisfying $(\pi_{1})_{*}\widehat{X}^{1} = X_{1}$ and $(\pi_{2})_{*}\widehat{X}_{1} = 0$. Literally correct articulations include \begin{align*} (\Ric_{1} \oplus \Ric_{2})(\widehat{X}_{1} + \widehat{X}_{2}, \widehat{Y}_{1} + \widehat{Y}_{2}) &= \Ric(\widehat{X}_{1} + \widehat{X}_{2}, \widehat{Y}_{1} + \widehat{Y}_{2}) \\ &= \Ric_{1}(X_{1}, Y_{1}) + \Ric(X_{2}, Y_{2}). \end{align*}
It's not entirely unreasonable to drop the hats (which gives your first alternative), but doing so is strictly an abuse of notation. Ultimately, you'll have to be guided (in a reader-dependent way) by the contrary constraints of (i) what gets the point across (ii) without ambiguity.