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When multiplying 2 square matrices the resulting first row ends up being a combination of the first column of a matrix and the first row of one another. Aren't we geometrically missing information?

Edit: What I mean is that to compute each composition of transformations there is a certain algorithm which leads to the correct result: 1.
I find it hard to ignore the way it is built up; why is it that to find the new image of i-hat (as shown in the picture) we multiply its first component by the x components (first column) of the second transform, and its second component by the y components? I at first thought we were just "saving" the motion/displacement exherted upon the first transform's "i-hat" components and that it didn't really matter in which component we saved it cause, well, it will end up being a single vector -combination of all the components. But I then discovered that the order followed was the unique path to get the correct composition. All the answers I've seen regarding this question state that this arises the "natural properties" of composition, but the matter is, why? There must be a geometrical explanation as to why we map the given vectors into the images of certain components and not otherwise.

I am deeply sorry if I am being unclear or if this is is just a stupid question that arises from a sequence of misconceptions about matrix multiplication, but it has been stoping me from going any further in linear algebra so I'd appreciate if someone told me what is really going on.

1.Let $E={e_1,...,e_n}$ be any vector-space basis for $Bbb R^n.$ For every $vinBbb R^n$ there is a unique sequence $v_1,...,v_n$ of numbers such that $v=v_1e_1+...+v_ne_n.$ And $v_j$ is called "the $j$-th co-ordinate of $v$ with respect to the basis $E$".

Note that the $j$-th co-ordinate of a sum of some $k$ vectors is the sum of the individual vectors' $j$-th co-ordinates. That is, $(,v(1)+...+v(k),)_j=v(1)_j
,+...+,v(k)_j;.$ And of course $(rv)_j=rcdot v_j$ for any $rin Bbb R.$

2. A linear map $f:Bbb R^nto Bbb R^n$ is uniquely and completely determined by $f(e_1),...,f(e_n).$ Because the co-ordinates $v_1,...,v_n$ are uniquely determined by a given $v,$ and $$f(v)=f(v_1e_1+..+v_ne_n)=f(v_1e_1)+...+f(v_ne_n)=v_1f(e_1)+...+v_n f(e_n).$$




3. Let $f_{j,i}$ be the $j$-th co-ordinate (with respect to $E$) of $f(e_i)$. For any $vin Bbb R^n,$ and any $j$, the $j$-th co-ordinate of $f(v)$ is $$ (f(v))_j=(,v_1f(e_1),+...+,v_n f(e_n),)_j =$$ $$=(v_1f(e_1))_j,+...+,(v_nf(e_n))_j=$$ $$=v_1f_{j,1},+...+,v_nf_{j,n};.$$

The array $(f_{j,i})_{1leq ileq n, 1leq jleq n}$ is called "the matrix representation of $f$ with respect to $E$".

Remark: Any $vin Bbb R^n$ is a sequence $(r_1,...,r_n)$ of numbers, and one choice for a basis $E$ is, for each $i$, to let $e_i$ be the sequence $(e_{i,1},,...,e_{i,n})$ in which $e_{i,i}=1$ and $e_{i,j}=0$ when $jne i$. With respect to this basis, if $v=(r_1,...,r_n)$ then $v_j=r_j.$

Matrix multiplication is feeding one linear change of variables into another. See my answer at Why, historically, do we multiply matrices as we do?.